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THE PENGUIN 0

BOOK OF

AND

PUZZLES
David Wells
PENGUIN BOOKS
THE PENGUIN BOOK OF
CURIOUS AND INTERESTING PUZZLES
David Wells was born in 1940. He had the rare distinction of
being a Cambridge scholar in mathematics and failing his degree.
He subsequently trained as a teacher and, after working on
computers and teaching machines, taught mathematics and
sCIence in a primary school and mathematics in secondary
schools. He is still involved with education through writing and
working with teachers.
While at university he became British under-21 chess champion,
and in the mIddle seventies was a game inventor, devising
'Guerilla' and 'Checkpoint Danger', a puzzle composer, and the
puzzle editor of Games & Puzzles magazine. From 1981 to
1983 he published The Problem Solver, a magazine of
mathematical problems for secondary pupils.
He has published several books of problems and popular
mathematics, including Can You Solve These? and Hidden
Connections, Double Meanmgs, and also Russia and England,
and the Transformations of European Culture. He has written
The Penguin Dictionary of Curious and Interesting Numbers and
The Penguin Dictionary of Curious and Interesting Geometry,
and is currently writing a book on the nature, learning and
teaching of mathematics.
David Wells
The Penguin Book of
Curious and Interesting Puzzles
PENGUIN BOOKS
PENGUIN BOOKS
Publish<-d by Ihe Penguin Group
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Firsl published 1992
10 9 8
Oavid Wells, 1992
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Contents
Acknowledgements VI
Introduction Vll
The Puzzles
The Solutions 179
Bibliography 373
Index 379
Acknowledgements
Please note that detailed sources for puzzles are given at the end of
each puzzle solution, where appropriate.
Grateful acknowledgement is given to the following:
Dover Publications, Inc., for permission to reproduce material from:
Stephen Barr, Second Miscellany of Puzzles (1969); A. H. Beiler,
Recreations in the Theory of Numbers (1966); Angela Dunn (ed.),
Mathematical Bafflers (1980), and The Second Book of Mathematical
Bafflers (1983); L. A. Graham, Ingenious Mathematical Problems and
Methods (1959), and The Surprise Attack in Mathematical Problems
(1968); J. A. H. Hunter, More Fun with Figures (1966); F. Mosteller,
Fifty Challenging Problems in Probability (1987); F. Schuh, The
Master Book of Mathematical RecreatIons (1968); George J. Summers,
New Puzzles in Logical Deduction (1968).
I will also note here that although the original Loyd and Dudeney
books are long out of print, two collections of Loyd's puzzles, both edited
by Martin Gardner, are pu blished by Dover under the titles Mathematical
Puzzles of Sam Loyd and More Mathematical Puzzles of Sam Loyd, and
they have also reprinted Dudeney's Amusements in Mathematics.
Robert Hale Ltd, for permission to reproduce 'Room for More
Inside', from Gyles Brandreth, The Complete Puzzler (1982).
McGraw-Hill, Inc., for permission to reproduce 'The True!', from
David Silverman, Your Move (1971).
Weidenfeld and Nicholson for permission to reproduce the Tangram
puzzles from E. Cuthwellis (ed.), Lewis Carroll's Bedside Book (1979).
John Hadley for the translation of Alcuin's Propositiones ad acuen-
dos juvenes, and David Singmaster for lending me his copy, as well as
giving me the run of his library of mathematical recreations. John
Hadley's complete translation has subsequently been published in the
Mathematical Gazette, Vol. 76, No. 475, March 1992.
Finally, I should like to thank the staff of the British Library for their
courteous help.
Introduction
To puzzle and be puzzled are enjoyable experiences, so it is no
surprise that puzzling problems are as old as history itself, and follow
a similar pattern of many centuries of slow progress, followed by
rapid expansion in the nineteenth century, and an explosion in the
twentieth. This book follows that pattern. The first third is devoted
to puzzles from the dawn of history, in Egypt and Babylon, up to the
nineteenth century. These are followed by examples of the puzzles of
Loyd and Dudeney, who straddle the nineteenth and twentieth centu-
ries, and other famous puzzlers of that era such as Lewis Carroll and
Eduard Lucas. The second half of the book is devoted to the great
variety of puzzles composed in the twentieth century.
I must emphasize, however, that this is not a history. I have merely
selected some representative figures. One day a history of puzzles will
be written, I hope by David Singmaster, who has spent many years
delving into the origins of popular puzzles, but in the meantime this
book will give readers examples, only, of the puzzling questions that
have found popular favour over the centuries.
Limitations of space have forced a strict selection. Word puzzles
are entirely excluded. I hope that in due course they will form a
separate volume, well justified by their immense richness and variety.
A boundary also had to be drawn between puzzles of a logical and
mathematical nature, and mathematical recreations and mathematics
itself. ThiS boundary cannot be drawn precisely, but generally speak-
ing problems which require any mathematics beyond the most elemen-
tary algebra and geometry, have been excluded, and few of the
puzzles require even that level of sophistication.
A number of puzzles are included which relate to mathematical
recreations or which led to the development of specific recreations,
but the recreations themselves are not treated. Readers interested in
mathematical recreations will find references to many of the best-
known and most readily available sources in the bibliography.
V11l Introduction
Finally, manipulative puzzles requiring some kind of apparatus
also deserve a book-length treatment of their own, and are excluded
here. All the puzzles in this book can be tackled either mentally, or
with the assistance of at most pencil and paper and perhaps a few
counters.
Compiling this book has taken me back to the days when I was
Puzzle Editor of Games & Puzzles magazine, and work was a pleasure
hard to distinguish from play. I hope that readers will find some of
that pleasure in the immense variety of puzzles assembled here.
I shall be happy to receive readers' opinions and suggestions,
though I cannot guarantee to respond to every letter personally. I
wish you happy and successful puzzling!
D.W. 1992
The Puzzles
The World's Oldest Puzzle
1. There are seven houses each containing seven cats. Each cat kills
seven mice and each mouse would have eaten seven ears of spelt.
Each ear of spelt would have produced seven hekats of grain. What is
the total of all these?
This puzzle, freely paraphrased here, is problem 79 in the Rhind
papyrus, our richest source for ancient Egyptian mathematics, which
is named after the Scottish Egyptologist A. Henry Rhind, who pur-
chased it in 1858 in Luxor.
The Rhind papyrus is in the form of a scroll about eighteen and a
half feet long and thirteen inches wide, written on both sides. It dates
from about 1650 BC. The scribe's name was Ahmes, and he states that
he is copying a work written two centuries earlier, so the original of
the Rhind papyrus was written in the same period as another famous
source of Egyptian mathematics, the Moscow papyrus, dating from
1850 BC.
Returning to the cats and mice, about 2800 years after Ahmes,
Fibonacci in his Liber Abaci (1202) posed this puzzle:
2. Seven old women are travelling to Rome, and each has seven
mules. On each mule there are seven sacks, in each sack there are
seven loaves of bread, in each loaf there are seven knives, and each
knife has seven sheaths. The question is to find the total of all of
them.
The resemblance is so strong that surely Fibonacci's problem is a
direct descendant, along an historical path that we can no longer
trace, of the Rhind puzzle? Not necessarily. There is an undoubted
fascination with geometrical series, and the number 7 is not only as
4 Penguin Book of Curious and Interesting Puzzles
magical and mysterious as any number can be, but was especially
easy for the Egyptians to handle, because they multiplied by repeated
doubling, and 7 = 1 + 2 + 4. Put these factors together, and you
naturally arrive at two similar puzzles.
The St Ives Riddle
3. As I was going to St Ives,
I met a man with seven wives.
Every wife had seven sacks,
Every sack had seven cats,
Every cat had seven kits;
Kits, cats, sacks and wives,
How many were going to St Ives?
This rhyme appears in the eighteenth-century Mother Goose collec-
tion. Is it also descended from the Rhind papyrus and Fibonacci?
Egyptian Fractions
The Egyptians could easily handle simple fractions, but with one
remarkable peculiarity. The only fractions they used were i and the
reciprocals of the integers, the so-called unit fractions with unit
numerators.
The Rhind papyrus contains a table of fractions in the form 21n for
all odd values of n from 5 to 101. They also had a rule for expressing
i of a unit fraction as the sum of unit fractions: to find i of t, multiply
5 by 2 and by 6: i of t = rl; + ~ . Similarly, i of i is -k + is. Curious
though this treatment of fractions may seem to us, no doubt it
seemed both natural and easy to them.
Thus their answer to the problem, 'divide seven loaves among ten
men' was not 7/10 of a loaf each, but the fraction t + t.
Can all proper fractions be expressed as the sum of unit fractions,
without repetition? Yes, as Fibonacci showed, also in his Liber Abaci,
where he described what is now called the greedy algorithm. Subtract
the largest possible unit fraction, then do the same again, and so on.
Sylvester proved in 1880 that applying this greedy algorithm to the
fraction plq, where p is less than q, produces a sequence of no more
than p unit fractions.
4. The greedy algorithm does not work so well if we add the
The Puzzles 5
condition that all the denominators must be odd. There are just five
ways to represent 1 as the sum of the smallest possible number of
Egyptian fractions, with odd denominators. Which has the smallest
largest denominator?
5. What is the smallest fraction 3/n for which the greedy algorithm
produces a sum in three terms, but two terms are actually sufficient?
The sum of the series 1 + 112' + 113
'
+ 114' ... = x'/6, so the sum
of different Egyptian fractions whose denominators are squares cannot
exceed x'/6, but might equal, for example, 1.
6. How can 1 be represented as the sum of unit fractions with square
denominators, with no denominator greater than 35
'
?
Think of a Number
7. Problem 29 of the Rhind papyrus is not quite so clear, but it is
plausibly the first ever 'Think of a Number' problem. It reads, 'Two-
thirds is to be added. One-third is to be subtracted. There remains
to.' In clearer language that reads: 'I think of a number, and add to it
two-thirds of the number. I then subtract one-third of the sum. My
answer is 10. What number did I think of?'
8. 'If the scribe says to thee, "10 has become j + to of what?'" is the
Egyptian way of saying, in effect, 'I think of a number. Two-thirds of
the number plus its tenth make to. What was the number?'
9. 'A number, plus its two-thirds, and plus its half, plus its seventh,
makes 37. What is ,the number?'
Readers will naturally wish to express the answer in Egyptian
fractions!
Sharing the Loaves
Arithmetic progressions have not been as popular in the history
of puzzles as geometric ones. There is after all something im-
pressive, mysterious even, in the rapidity with which geometric
progressions increase, while arithmetic progressions just plod along,
step by equal step.
Yet puzzles about arithmetical pr->gressions can be thought-
provoking, as this example illustrates.
6 Penguin Book of Curious and Interesting Puzzles
10. 'A hundred loaves to five men, one-seventh of the three first men
to the two last.'
The meaning is: 'Divide 100 loaves between five men so that the
shares are in arithmetical progression, and the sum of the two smaller
shares is one-seventh of the sum of the three greatest.'
Squares Without Pythagoras
It is a well-known 'fact' that the ancient Egyptians used knotted ropes
to make a 3-4-5 triangle and hence construct accurate right-angles.
This 'fact' is actually a myth, based on a suggestion by the historian
Moritz Cantor that the Egyptians might just possibly have made
right-angles this way. There is no evidence that they did anything of
the sort, or that they had any knowledge whatsoever of Pythagoras's
theorem. They did, however, consider problems about areas and
square numbers. This is from the Berlin papyrus:
11. 'If it is said to thee ... the area of a square of 100 is equal to that
of two smaller squares. The side of one is f + i the side of the other.
Let me know the sides of the two unknown squares.'
The Babylonians
Babylonian mathematics was arithmetical and algebraic and far in
advance of Egyptian mathematics of the same period. They could
solve all the problems in the Rhind papyrus and many more besides.
The Babylonians counted in a sexagesimal system. Instead of
counting in tens and hundreds and using tenths and hundredths, and
so on, they used multiples of 60, so 6,30 means 6 + (30/60), or 61,
and 11,22,30 means 11 + (22/60) + (30/3600), or 11i.
Dividing a Field
12. A triangular field is to be divided between six brothers by
equidistant lines parallel to one side. The length of the marked side is
6,30 and the area is 11,22,30. What is the difference between the
brothers' shares?
This problem is much like Problem to, which required the construc-
tion of an arithmetical series to fit given conditions. Other problems
were far more advanced. Thus a tablet from about 1600 Be, contempor-
The Puzzles 7
ary with the Rhind papyrus, leads in modern notation to the solution
of two equations of the form:
bx' cy'
xy=a -+-+d=O
y x
which leads to an equation in x·, x' and a constant.
13. This is from about 1800 Be:
'An area A, consisting of the sum of two squares, is 1000. The side
of one square is 10 less than two-thirds of the other square. What are
the sides of the squares?'
Pythagorean Triples
The Babylonians, unlike the Egyptians, not only knew Pythagoras's
theorem, but they were also familiar with Pythagorean triples, triples
of whole numbers such as 3-4-5 which are the sides of right-angled
triangles. Their investigations of Pythagorean triples started a trail of
discovery, leading through Diophantus to Fermat, to the present day.
14. Ladders were a natural source of problems. A ladder of length
0,30 is standing upright against a wall. If the upper end slides down
the wall a distance of 0,6, how far will the lower end move out from
the wall?
'Plimpton 322' is the name of a clay tablet dating from between 1900 Be
and 1600 Be. It contains fifteen numbered lines with two figures in each
line which are the hypotenuse and one leg of a right-angled triangle.
Although the lengths given seem to vary in an apparently irregular
way from one line to the next, in fact their ratios increase steadily
from 169/119 = 1.42 in the first line to 106/56 = 1.89 in the last.
15. Problem: find the hypotenuse and one leg of a right-angled
triangle whose ratio is approximately 1.54.
The Greeks
Archimedes' Cattle Problem
Archimedes (287-212 Be) was the greatest mathematician of antiquity,
a wonderful geometer who anticipated the calculus, invented hydro-
8 Penguin Book of Curious and Interesting Puzzles
statics, and studied giant numbers in his book The Sandreckoner. It is
a curiosity that one extremely difficult problem and one simple
recreation are associated with his name.
16. 'If thou art diligent and wise, 0 stranger, compute the number of
cattle of the Sun, who once upon a time grazed on the fields of the
Thrinician isle of Sicily, divided into four herds of different colours,
one milk white, another glossy black', the third yellow and ,the last
dappled. In each herd were bulls, mighty in number according to
these proportions: understand, stranger, that the white bulls were
equal to a half and a third of the black together with the whole of the
yellow, while the black were equal to the fourth part of the dappled
and a fifth, together with, once more, the whole of the yellow.
Observe further that the remaining bulls, the dappled, were equal to a
sixth part of the white and a seventh, together with all the yellow.
These were the proportions of the cows: the white were precisely
equal to the third part and a fourth of the whole herd of the black;
while the black were equal to the fourth part once more of the
dappled and with it a fifth part, when all, including the bulls, went to
pasture together. Now the dappled in four parts were equal in
number to a fifth part and a sixth of the yellow herd. Finally the
yellow were in number equal to a sixth part and seventh of the white
herd. If thou canst accurately tell, 0 stranger, the number of cattle of
the Sun, giving separately the of well-fed bulls and again the
number of females according to each colour, thou wouldst not be
called unskilled or ignorant of numbers, but not yet shalt thou be
numbered among the wise ...
'But come, understand also all these conditions the cows
of the Sun. When the white bulls mingled their number with the
black, they stood firm, equal in depth and breadth, and the plains of
Thrinacia, stretching far in all ways, were filled with their multitude.
Again, when the yellow and the dappled bulls were gathered into one
herd they stood in such a manner that their number, beginning from
one, grew slowly greater till it completed a triangular figure, there
being no bulls of other colours in their midst nor none of them
lacking.
'If thou art able, 0 stranger, to find out all these things and gather
them together in your mind, giving all the relations, thou shalt depart
crowned with glory and knowing that thou hast been adjudged
perfect in this species of wisdom.'
Archimedes' cattle problem is extant in more than one manuscript.
The Puzzles 9
The 'most complete' version contains the extra conditions that follow
the ellipsis. These conditions are ambiguous: because the bulls are
longer than they are broad, the condition that the white and black
bulls together form a square does not necessarily mean that their total
is a square number; it could be merely a rectangular number.
It is plausible that the more difficult interpretation is intended.
Archimedes dedicated the problem to his friend the great Alexandrian
astronomer Eratosthenes, which suggests that it was extremely diffi-
cult, and Archimedes' interest in very large numbers is evident from
his Sandreckoner, in which he calculated the number of grains of
sand needed to fill a sphere whose centre was the centre of the earth
and which extended to reach the sun. Also, in classical antiquity a
difficult problem was often described as a problema bovinum or a
problema Archimedis, such was his fame. If this is so, then the
solution is indeed complex and extraordinarily lengthy. A. Amthor
calculated in 1880 that the total number of cattle in this case is a
number of 206,545 digits. Further details will be found in Sir Thomas
Heath's A History of Greek Mathematics, p. 319.
If, however, the latter conditions are ignored, and the reader is
willing to be judged merely 'not unskilled' in the art, rather than
perfectly wise, then the answer will be found in the Solutions section.
Loculus of Archimedes
Several ancient sources refer to this puzzle, which is described in an
Arabic manuscript, The Book of Archimedes on the Division of the
Figure Stomaschion.
10 Penguin Book of Curious and Interesting Puzzles
The Loculus consists of fourteen pieces making a square. The
method of division is almost self-evident: M, T and C are mid-points,
and HK passes through A and OC through B.
The object of the puzzle is to make figures with these pieces.
Unlike the Chinese Tangram puzzle, which might be said to have too
few pieces, this has rather a lot. (Is it a coincidence that the Tangram
has the magical number of seven pieces and the Loculus exactly twice
as many?)
17. How can this figure of an elephant be composed from the pieces
of the Loculus?
Light Reflected off a Mirror
A ray of light passes from point A to point B, by bouncing off the
surface of a plane mirror. Assuming that light always travels by the
shortest path, where does it strike the mirror?
This beautiful and important problem occurs in the Catoptrica of
Heron of Alexandria (c. 75 AD). Heron's assumption is correct, so it
has important practical applications.
A modern version is the following:
18. Mary, who is standing at S, wishes to walk to the river for a
drink and then back to T, walking as short a distance as possible. To
what point on the river bank should she walk?
The Puzzles 11
-s
19. From the Greek Anthology, c. 500 AD: 'I am a brazen lion; my
spouts are my two eyes, my mouth and the flat of my right foot. My
right eye fills a jar in two days, my left eye in three, and my foot in
four. My mouth is capable of filling it in six hours; tell me how long
all four together will take to fill it?'
Heron was a master of mechanical devices. His Pneumatica describes
scores of machines operated by wind and water, so it is no surprise
that the famous cistern problem occurs in his Metrika.
Famous? This is the infamous problem about the tank which is
filled with water from several pipes, which was still being used to
torture schoolchildren till the middle of this century, and which has
become a byword for 'useless' mathematics. This is a great pity,
because the idea behind it is far from useless and turns up in many
important situations.
20. From The Tutorial Arithmetic by W. P. Workman, published in
1920: 'A and B together can do a piece of work in 6 days, Band C
together in 20 days, C and A together in 7t days. How long will each
require separately to do the same work?'
Heron was also a geometrician:
21. Find two rectangles, with integral sides, such that the area of the
first is three times the area of the second, and the perimeter of the
second is three times the perimeter of the first.
12 Penguin Book of Curious and Interesting Puzzles
22. In a right-angled triangle with integral sides, the sum of the area
and the perimeter is 280. Find the sides and the area.
The First Pure Number Puzzles
23. What number must be added to 100 and to 20 (the same number
to each) so that the sums are in the ratio 3:1?
24. Two numbers are such that if the first receives 30 from the
second, they are in the ratio 2:1, but if the second receives 50 from the
first, their ratio is then 1:3. What are the numbers?
25. The sums of four numbers, omitting each of the numbers in turn,
are 22, 24, 27 and 20, respectively. What are the numbers?
These problems are nos. 8, 15 and 17 of Book I of the Arithmetica of
Diophantos of Alexandria (c. 250 AD). Typically, the solutions are all
whole numbers. All his problems concern integers or rational numbers,
and such problems in integers are named Diophantine after him.
While studying such problems he is led to discuss the multiplication of
positive and negative numbers. Coincidentally, a commentary on his
work was written by Hypatia (c. 410), the first known woman math-
ematician, who was murdered by a Christian mob in the year 415.
The works of Diophantos vary from the simply puzzling and
puzzlingly simple, to very difficult questions which had a stunning
impact when his works were first translated into Latin and studied by
European mathematicians more than 1200 years later. Xylander wrote
in 1575: 'I came to believe that in Arithmetic and Logistic "I was
somebody". And in fact by not a few, among them some true
scholars, I was adjudged an Arithmetician beyond the common order.
But when I first came upon the work of Diophantos, his method and
his reasoning so overwhelmed me that I scarcely knew whether to
think of my former self with pity or with laughter.'
The elementary problems that Diophantos solves could all have
been presented, had he so wished, as puzzles in everyday settings, and
were by other writers. Here the numbers themselves are personified:
26. 'To find three numbers such that, if each give to the next
following a given fraction of itself, in order, the results after each has
given and taken may be equal.
'Let the first give! of itself to the second, the second give i of itself
to the third, and the third give t of itself to the first. What are the
The Puzzles 13
numbers?' (Diophantos assumes that all these transactions take place
simultaneously, and not in sequence.)
Square Problems
27. Find three numbers such that the product of any two added to
the third gives a square.
28. Find three numbers such that their sum is a square and the sum
of any pair is a square.
29. 'A man buys a certain number of measures of wine, some at 8
drachmas, some at 5 drachmas each. He pays for them a square
number of drachmas; and if we add 60 to this number, the result is a
square, the side of which is equal to the whole number of measures.
Find how many he bought at each price.'
The Area Enclosed Against the Seashore
So they reached the place where you will now behold mighty
walls and the rising towers of the new town of Carthage; and
they bought a plot of ground named Byrsa ... for they were to
have as much as they could enclose with a bull's hide.
Virgil, Aeneid, Book I, II. 360-70
Questions and facts about extremes have a natural attraction;
witness the runaway success of the Guinness Book of Records. At a
more serious level, many scientific principles can be expressed in
terms of maxima and minima, as Heron's problem of the ray of light
reflecting off a mirror illustrates (see p. 10).
30. 'Given a long string, with which to enclose the maximum possible
area against a straight shore-line, how should the string be disposed?'
Here are two variants:
14 Penguin Book of Curious and Interesting Puzzles
31. This frame is composed of four rods that ar'e hinged to each other
at their ends. When will the area enclosed by the frame be a maximum?
32. This figure shows the corner of a room with a screen, composed
of two identical halves hinged together, placed to cut off a portion of
the corner of the room. How should the screen be placed to enclose
as large an area as possible?
33. An isosceles triangle has two equal sides of length 10, hinged
together. What is the maximum area of the triangle?
Metrodorus and the Greek Anthology
The Greek Anthology is a collection of literary verses and epigrams.
Surprisingly, Book XIV comprises a large number of riddles, enigmas
and puzzles, credited to Metrodorus (c. 500 AD).
These include not only arithmetical puzzles, but very early word
puzzles, including beheadings, in which a word loses letter after letter
from its front end but always remains a proper word, and this puzzle:
If you put one hundred in the middle of a burning fire, you will
find the son and a slayer of a virgin.
The answer is to put the Greek symbol for 100, rho, into the word
for fire, pyros, to get Pyrrhos, the son of Deidamia and the slayer of
Polyxena.
The Puzzles 15
The arithmetical and logical puzzles include what were already
classic problems, such as finding the weights of bowls given in
arithmetical progression, and the cisterns problem, and new types:
My father-in-law killed my husband and my husband killed my
father-in-law; my brother-in-law killed my father-in-law, and
my father-in-law my father.
The answer is Andromache. Achilles, father of her second husband,
Pyrrhus, killed Hector, Pyrrhus killed Priam, Paris killed Achilles, and
Achilles killed her father, Eetion.
34. '''Best of clocks, how much of the day is past?" There remains
twice two-thirds of what is gone.' (Problem 6; the day is counted as
lasting for 12 hours.)
35. This tomb holds Diophantos. Ah, how great a marvel! the tomb
tells scientifically the measure of his life. God granted him to be a boy
for the sixth part of his life, and adding a twelfth part to this, He
clothed his cheeks with down; He lit him the light of wedlock after a
seventh part, and five years after his marriage He granted him a son.
Alas! late-born wretched child; after attaining the measure of half his
father's life, chill Fate took him. After consoling his grief by this
science of numbers for four years he ended his life.' (Problem 126)
36. 'I desire my two sons to receive the thousand staters of which I
am possessed, but let the fifth part of the legitimate one's share
exceed by ten the fourth part of what falls to the illegitimate one.'
(Problem 11)
Arabic Puzzles
AI-Khwarizmi (c. 825 AD)
Al-Khwarizmi wrote a book, al-Kitab al-mukhtasar hisab al-jabr
wa'l-muqabala, or The Compendious Book on Calculations by Com-
pletion and Balancing, on the solution of equations. Later Arabic
works tended to use the same expression al-jabr wa'l-muqabala, or
just al-jabr, to refer to books on the same theme, from whence we
eventually derive our word 'algebra'.
The second half of the same book deals with problems of inherit-
ance, according to Islamic law. This is an essential study for lslamic
16 Penguin Book of Curious and Interesting Puzzles
jurists, as it had previously been for Roman lawyers, though as Ibn
Khaldun wrote in the fourteenth century, 'Some authors are inclined
to exaggerate the mathematical side of the discipline and to pose
problems requiring for their solution various branches of arithmetic,
such as algebra, the use of roots, and similar things' (Berggren, 1986,
p.53).
Well, mathematicians would, wouldn't they! This problem is
practical:
37. A woman dies, leaving her husband, a son and three daughters.
She also leaves t + j of her estate to a stranger. According to law, the
husband receives one quarter of the ~ t a t e and the son receives double
the share of a daughter, but this division is made only after the legacy
to the stranger has been paid. How must the inheritance be divided?
Abu) Wafa (940-998)
Abul Wafa was born in Buzjan in Persia in 940. He wrote comment-
aries on Euclid and Diophantos and AI-Khwarizmi, but he is best
known for his study of geometrical dissections and of constructions
with a rusty compass, meaning a compass which is so stiff that it can
be used with only one opening.
38. Construct an equilateral triangle inside a square, so that one
vertex is at a corner of the square and the other two vertices are on
the opposite sides.
39. Three Squares into One Dissect three equal squares into one
square.
40. Dissect two identical larger squares plus one smaller square into
one square.
41. How can two regular hexagons, of different sizes, be dissected
into seven pieces which fit together to make one, larger, regular hexa-
gon?
42. Given three identical triangles, and one smaller triangle similar to
them in shape, how can all four be dissected into one triangle?
43. The Rusty Compass Using only a straight-edge and a compass
with a fixed opening, construct at the endpoint A of a segment AB a
The Puzzles 17
perpendicular to that segment, without prolonging the segment
beyond A.
44. Using only a straight-edge and fixed-opening compasses, divide a
given line-segment into any given number of equal parts.
45. Construct a regular pentagon in a given circle, using only a
straight-edge and a compass with a fixed opening ~ q u a l to the radius
of the circle.
Sissa and the Chessboard
Ibn Kallikan (c. 1256) was the first author to tell the story of Sissa
ben Dahir, who was asked by the Indian King Shirham what he
desired as a reward for inventing the game of chess:
46. '''Majesty, give me a grain of wheat to place on the first square,
and two grains of wheat to place on the second square, and four
grains of wheat to place on the third, and eight grains of wheat to
place on the fourth, and so, Oh King, let me cover each of the sixty-
four squares on the board."
'''And is that all you wish, Sissa, you fool?" exclaimed the aston-
ished King.
'''Oh, Sire," Sissa replied, "I have asked for more wheat than you
have in your entire kingdom, nay, for more wheat than there is in the
whole world, verily, for enough to cover the whole surface of the
earth to the depth of the twentieth part of a cubit.'"
How many grains of wheat did Sissa require?
Indian Puzzles
The Bhakshali manuscript was found in 1881 in north-west India and
dates from somewhere between the third and twelfth centuries, depend-
ing on which authority you choose. It contains the earliest - if it
really dates as early as the third century - version of what came to be
called 'One Hundred Fowls' problem (see Problem 74), in this form:
47. Twenty men, women and children earn twenty coins between
them. Each man earns 3 coins, each woman It coms and each child t
coin. How many men, women and children are there?
18 Penguin Book of Curious and Interesting Puzzles
Mahavira (c. 850) wrote on elementary mathematics. Problems 48 to
54 are from his book the Ganita-Sara-Sangraha.
48. 'Three puranas formed the pay of one man who is a mounted
soldier; and at that rate there were sixty-five men in all. Some (among
them) broke down, and the amount of their pay was given to those
that remained in the field. Of this, each man obtained 10 puranas.
You tell me, after thinking well, how many remained in the field and
how many broke down.'
49. 'Two market-women were selling apples, one at two for 1 cent,
and the other at three for 2 cents. They had thirty apples apiece. In
order to end their competition they formed a trust, pooling their
stock and selling the apples at five for 3 cents. This was to their
advantage, since under the new arrangement they took, in total, 36
cents, while under the old system they would have received a total of
only 35 cents.
'Two other women, who also had thirty apples apiece, and who
were selling them at two for 1 cent and three for 1 cent, also formed a
trust to sell their apples, at five for 2 cents. But instead of the total of
25 cents which they would have taken in operating separate enter-
prises, their trust grossed only 24 cents. Why?'
50. 'One night, in a month of the spring season, a certain young lady
... was lovingly happy along with her husband on ... the floor of a
big mansion, white like the moon, and situated in a pleasure-garden
with trees bent down with the load of bunches of flowers and fruits,
and resonant with the sweet sounds of parrots, cuckoos and bees
which were all intoxicated with the honey obtained from the flowers
therein. Then on a love-quarrel arising between the husband and the
wife, that lady's necklace made up of pearls became sundered and fell
on the floor. One-third of that necklace of pearls reached the maid-
servant there; one-sixth fell on the bed; then one-half of what remained
(and one-half of what remained thereafter and again one-half of what
remained thereafter) and so on, counting six times [in all] fell all of
them everywhere; and there were found to remain [unscattered] 1,161
pearls; and if you know ... give out the measure of the pearls.'
51. 'In how many ways can different numbers of flavours be used in
combination together, being selected from the astringent, the bitter,
the sour, the pungent, and the saline, together with the sweet taste?'
The Puzzles 19
52. 'Three merchants saw in the road a purse [containing money].
One said, "If I secure this purse, I shall become twice as rich as both
of you together."
'Then the second said, "I shall become three times as rich."
'Then the third said, "I shall become five times as rich."
'What is the value of the money in the purse, as also the money on
hand [with each of the three merchants]?'
53. Arrows, if they are thin cylinders, circular in cross-section, can be
packed in hexagonal bundles:
'The circumferential arrows are eighteen in number. How many [in
all] are the arrows to be found [in the bundle] within the quiver?'
54. Two pillars are of known height. Two strings are tied, one to the
top of each. Each of these two strings is stretched so as to touch the
foot of the other pillar. From the point where the two strings meet,
another string is suspended vertically till it touches the ground. What
is the length of this suspended string?
This is identical to puzzles about ladders resting across passageways
in which the heights of the points at which they touch are given. If
not the vertical heights bur the lengths of the ladders are known, then
the problem of finding the height of their intersection is far harder.
(See p. 131.)
Bhaskara (IllS-c. l18S) was an astronomer and mathematician whose
most famous work, the Lilavati, from which the following problems
20 Penguin Book of Curious and Interesting Puzzles
are taken, was addressed to his daughter, or perhaps his wife. It ends
with this delightful paragraph, typical of the Indian style of the
period:
Joy and happiness is indeed ever increasing in this world for
those who have Lilavati clasped to their throats, decorated as
the members are with neat reduction of fractions, multiplication
and involution, pure and perfect as are the solutions, and tasteful
as is the speech which is exemplified.
55. 'In an expedition to seize his enemy's elephants, a king marched 2
yojanas the first day. Say, intelligent calculator, with what increasing
rate of daily march did he proceed, since he reached his foe's city, a
distance of 80 yojanas, in a week?'
56. 'A snake's hole is at the foot of a pillar which is 15 cubits high
and a peacock is perched on its summit. Seeing a snake, at a distance
of thrice the pillar's height, gliding towards his hole, he pounces
obliquely upon him. Say quickly at how many cubits from the snake's
hole do they meet, both proceeding an equal distance?'
It is natural that Hindu writers should have considered sooner or
later the permutations and combinations of the attributes of their
gods:
57. 'How many are the variations in the form of the God Siva by the
exchange of his ten attributes held reciprocally in his several hands:
namely, the rope, the elephant's hook, the serpent, the tabor, the
skull, the trident, the bedstead, the dagger, the arrow, and the bow:
as those of Vishnu by the exchange of the mace, the discus, and lotus
and the conch?'
The final Hindu problem is unattributed, but on a popular theme:
58. The first man has sixteen azure-blue gems, the second has ten
emeralds, and the third has eight diamonds. Each among them gives
to each of the others two gems of the kind owned by himself; and
then all three men come to be possessed of equal wealth. What are
the prices of those azure-blue gems, emeralds and diamonds?
The Puzzles 21
Puzzles from China
The First Magic Square
59. How can the numbers 1 to 9 be arranged in the cells of this
square so that the sums of every row and column and both diagonals
are equal?
The resulting figure has essentially the arrangement of the Lo Shu,
which, in Chinese legend going back at least to the fifth century BC,
was the gift of a turtle from the River Lo to the Emperor Yu the
Great, who first controlled the flow of the Lo and the Yellow rivers.
The Nine Chapters
The Nine Chapters of Mathematical Art is supposed to have been
written in the third century BC, and contains the first known exam-
ples of the solution of linear simultaneous equations, well ahead of
the West, as well as the extraction of square and cube roots.
60. 'Suppose that there are a number of rabbits and pheasants
confined in a cage, in all thirty-five heads and ninety-four feet;
required the number of each?'
61. 'A number of men bought a number of articles, neither of which
are known; it is only known that if each man paid 8 cash, there
would be a surplus of 3 cash; and if each man paid 7 cash, there
would be a deficiency of 4 cash. Required the respective numbers?'
22 Penguin Book of Curious and Interesting Puzzles
62. 'If five oxen and two sheep cost 10 taels of gold, and two oxen
and five sheep cost 8 taels, what are the prices of the oxen and sheep
respectively?'
63. 'There are three classes of corn, of which three bundles of the
first class, two of the second class and one of the third make 39
measures. Two of the first, three of the second and one of the third
make 34 measures. And one of the first, two of the second and three
of the third make 26 measures. How many measures of grain are
contained in one bundle of each class?'
The following puzzles are from the ninth and last section of the book,
and all concern right-angled triangles and the Gougu theorem, as the
Chinese called what we call Pythagoras's theorem.
In contrast to later problems in Diophantos, these are all set in
remarkably realistic contexts, realistic that is if a mathematician
happened to notice a reed breaking the surface of a pool, or a chain
hanging from a pillar.
64. 'There is a pool 10 feet square, with a reed growing vertically in
the centre, its roots at the bottom of the pool, which rises a foot
above the surface; when drawn towards the shore it reaches exactly
to the brink of the pool; what is the depth of the water?'
65. 'A chain suspended from an upright post has a length of 2 feet
lying on the ground, and on being drawn out to its full length, so as
just to touch the ground, the end is found to be 8 feet from the post;
what is the length of the chain?'
The following problem was also presented by the Indian mathemati-
cian and astronomer Brahmagupta, more than 600 years later:
66. 'There is a bamboo 10 feet high, the upper end of which being
broken down on reaching the ground, the tip is just 3 feet from the
stem; what is the height of the break?'
67. 'What is the largest circle that can be inscribed wlthm a right-
angled triangle, the two short sides of which are respectively 8 and
IS?'
68. 'Of two water weeds, one grows 3 feet and the other 1 foot on
the first day. The growth of the first becomes every day half of that
The Puzzles 23
of the preceding day, while the other grows twice as much as on the
day before. In how many days will the two grow to equal heights?'
Sun Tsu Suan-Ching (fourth century AD)
69. 'A woman was washing dishes in a river, when an official whose
business was overseeing the waters demanded of her: "Why are there
so many dishes here?"
'''Because a feasting was entertained in the house," the woman
replied. Thereupon the official inquired the number of guests.
'''I don't know," the woman said, "how many guests there had
been; but every two used a dish for rice between them; every three a
dish for broth; every four a dish for meat; and there were sixty-five
dishes in all." ,
The next problem is an example of the famous Chinese Remainder
Theorem. Such problems had practical applications to calendar prob-
lems, when cycles of different lengths are compared.
70. 'There are certam things whose number is unknown. Repeatedly
divided by 3, the remainder is 2; by 5 the remainder is 3; and by 7 the
remainder is 2. What will be the number?'
71. 'There are three sisters, of whom the eldest comes home once
every five days, the middle in every four days, and the youngest in
every three days. In how many days will all the three meet together?'
Liu Hui (263 AD), in the Hai Tao Suan-Ching, or Sea-Island Arithmeti-
cal Classic, poses this simple puzzle:
72. What is the size of a square inscribed in the corner of a right-
angled triangle to touch the hypotenuse?
The Chang Sh'/u-Chien Suan-Ching, or The Arithmetical Classic of
Ch-iu Chien (sixth century), poses one of the earliest chasing and
returning puzzles:
73. 'A man, who had stolen a horse, rode away on its back. When he
had gone 37 miles, the owner discovered the theft and pursued the
thief for 145 miles; he then returned, [believing himself] unable to
overtake him. When he turned back the thief was riding 23 miles
ahead of him; If he had continued in his pursuit without coming
back, in how many further miles would he have overtaken him?'
24 Penguin Book of Curious and Interesting Puzzles
It also contains the earliest '100 fowls' problem:
74. 100 fowls are sold for 100 shillings, the cocks being sold for 5
shillings each, the hens for 3 shillings and the chicks for :1 shilling
each. How many of each were sold?
Yang Hui (c. 1270 AD) wrote an 'Arithmetic in Nine Sections', which
contains the very first extant representation of what we in the West
call Pascal's Triangle (from an earlier Chinese source, c. 1000 AD).
His book was called, apparently, Hsu Ku Chai Chi Suan Fa (1275). It
contains the following magic configuration:
75. Arrange the numbers 1 to 33 in these circles so that every circle
and eve"ry diameter has the same total.
'Propositions to Sharpen Up the Young'
Propositiones ad acuendos juvenes was written in the monastery of
Augsberg, about the year 1000, and has been included in the works of
Alcuin (c. 732-804), the English scholar and churchman who spent
his life at the court of the Emperor Charlemagne, on the grounds that
Alcuin writes in one of his letters to the Emperor that he is sending
him, among other matters, 'certain subtle figures of arithmetic, for
pleasure', and this might be that collection.
Anyway, it is the earliest European collection of mathematical and
The Puzzles 25
logical puzzles, and contains the first appearance of many well-known
puzzle types.
Like the problems of Metrodorus in the Greek Anthology, a few
riddles and trick questions find their way into the fifty-three problems
in the collection.
76. 'An ox ploughs a field all day. How many footprints does he
leave in the last furrow?' (Problem XIV)
'A man has 300 pigs, and orders that the pigs must be killed, an odd
number each day, in three days. Say how many pigs must be killed
each day.'
This is Problem XLIII. The answer is: 'This is a fable. Nobody can
solve how to kill 300 or 30 pigs in three days, an odd number each
day. This puzzle is given to children to solve.'
This could be cruelty to little children, but it is also an early
recognition that some problems simply cannot be solved.
77. 'Two wholesalers with 100 shillings between them bought some
pigs with the money. They bought at the rate of five pigs for 2
shillings, intending to fatten them up and sell them again, making a
profit. But when they found that it was not the right time of year for
fattening pigs, and they were not able to feed them through the
winter, they tried to sell them again to make a profit. But they
couldn't, because they could only sell them for the price they had
paid for them ... When they saw this, they said to each other: "let's
divide them". By dividing them, and selling them at the rate they had
bought them for, they made a profit. How many pigs were there, and
how could they be divided to make a profit, which could not be made
by selling them all at once?' (Problem VI)
78. 'A kmg ordered his servant to collect an army from thirty
manors, in such a way that from each manor he would take the same
number of men as he had collected up to then. The servant went to
the first manor alone; to the second he went with one other .. .' How
many men were collected in all? (Problem XIII)
79. 'If two men each take the other's sister in marriage, what is the
relationship between their sons?' (Problem XI)
80. 'A father, when dying, gave to his sons thirty glass flasks, of
which ten were full of wine, ten were half full, and the last ten were
26 Penguin Book of Curious and Interesting Puzzles
empty. Divide the [wine] and the flasks, so that each of the three sons
receives equally of both glass and wine.' (Problem XII)
The next three classic problems, like the last two, appear for the first
time in Alcuin.
81. Three Friends and their Sisters 'Three men, each with a sister,
needed to cross a river. Each one of them coveted the sister of
another. At the river, they found only a small boat, in which only two
of them could cross at a time. How did they cross the river, without
any of the women being defiled by the men?' (Problem XVII)
82. A Man, a Goat, and a Wolf 'A man takes a wolf, a goat and a
cabbage across the river. The only boat he could find could take only
two of them at a time. But he had been ordered to transfer all of these
to the other side in good condition. How could this be done?'
(Problem XVIII)
83. A Very Heavy Man and Woman 'A man and a woman, each
the weight of a loaded cart, with two children who between them
weigh as much as a loaded cart, have to cross a river. They find a
boat which can only take one cartload. Make the transfer, if you can,
without sinking the boat.' (Problem XIX)
84. 'A dying man left 960 shillings and a pregnant wife. He directed
that if a boy was born, he should receive three-quarters of the whole,
and the child's mother should receive one quarter. But if a daughter
was born, she would receive seven-twelfths, and her mother five-
twelfths. It happened however that twins were born - a boy and a
girl. How much should the mother receive, how much the son, and
how much the daughter?' (Problem XXV)
85. 'A stairway consists of 100 steps. On the first step stands a
pigeon; on the second, two pigeons; on the third, three; on the fourth,
four; on the fifth, five; and so on every step up to the hundredth.
How many pigeons are there altogether?' (Problem XLII)
The Puzzles 27
Liber Abaci
Leonardo of Pisa (c. 1175-1250) was a member of the Bonacci family,
so was often know., as Fibonacci (!ilio Bonacci). As a young man he
travelled to Bugia in North Africa to help h i ~ father, who directed a
trading post there, and learnt from the local Arabs the new Indian
numerals, our Hindu numerals, which he helped to introduce to
Europe.
He wrote a book on calculations, the Liber Abaci, a compendium
on geometry and trigonometry, Practica geometriae, and the Liber
quadratorum (The Book of Squares) on Diophantine problems.
He describes in the prologue to The Book of Squares how he was
invited to the court of Emperor Frederick II of Sicily to compete in a
mathematical tournament. He solved all three problems posed to him
by John of Palermo. The first, in the style of Diophantos, was to:
86. 'Find a rational number such that 5 added to, or subtracted from,
its square, is also a square.'
The second was to solve the cubic
Xl + 2x
2
+ lOx = 20
Leonardo found the solution, 1.3688081075, which is correct to nine
decimal places.
This is the third problem:
87. 'Three men possess a pile of money, their shares being 1/2, 1/3,
116. Each man takes some money from the pile until nothing is left.
The first man returns 112 of what he took, the second 1/3 and the
third 116. When the total so returned is divided equally among the
men it is found that each then possesses what he is entitled to. How
much money was in the original pile, and how much did each man
take from the pile?'
Breeding Rabbits
Fibonacci is best remembered for the following problem, which leads
to the Fibonacci sequence:
88. 'A certain man put a pair of rabbits in a place surrounded on all
sides by a wall. How many pairs of rabbits can be ,produced from
28 Penguin Book of Curious and Interesting Puzzles
that pair in a year if it is supposed that every month each pair begets
a new pair which from the second month on becomes productive?'
89. 'A lion would take four hours to eat one sheep; a leopard would
take five hours; and a bear would take six; we are asked, if a single
sheep were to be thrown to them, how many hours would they take
to devour it?'
90. 'A man left to his oldest son one bezant and a seventh of what
was left; then, from the remainder, to his next son he left two bezants
and a seventh of what was left; then, from the new remainder, to his
third son he left three bezants and a seventh of what was left. He
continued in this way, giving each son one bezant more than the
previous son and a seventh of what remained. By this division it
developed that the last son received all that was left and all the sons
shared equally. How many sons were there and how large was the
man's estate?'
91. 'A man entered an orchard [with] seven gates, and there took a
certain number of apples. When he left the orchard he gave the first
guard half the apples and one apple more. To the second guard he
gave one half of his remaining apples and one apple more. He did the
same to each of the remaining five guards, and left the orchard with
one apple. How many apples did he gather in the orchard?'
•
92. Serpent Climbing out of a Well Dell'Abaco (c. 1370) discusses
this famous puzzle. A serpent lies at the bottom of a well whose depth
is 30. It starts to climb, rising up 1 every day and falling back t at
night. How long does it take to climb out of the well?
93. The Best View of a Statue From what distance will a statue on
a plinth subtend the largest angle? (See figure opposite.)
If you are too close, the statue will appear greatly foreshortened,
but if you walk back too far, it will just appear small.
This problem was originally posed by Regiomontanus (1436-76) in
1471 to Christian Roder, as a question about a suspended vertical
rod. It is notable as the first extremal problem since the days of
antiquity and Heron's problem about the ray of light bouncing off a
mirror.
The Puzzles 29
The same problem has been re-invented many times, most recently
in this practical form:
According to the rules of rugby union football, a conversion of a
try must be taken on a line extending backwards from the point of
touchdown, at right-angles to the goal-line. From which point on this
line should the conversion be taken, if the aim is to maximize the
angle subtended by the goal-posts? This problem applies only when
the try is not scored between the posts.
The Couriers Meeting
The Treviso Arithmetic (1478) poses this problem:
94. 'The Holy Father sent a courier from Rome to Venice, command-
ing him that he reach Venice in seven days. And the most illustrious
Signoria of Venice also sent another courier to Rome, who should
reach Rome in nine days. And from Rome to Venice is 250 miles. It
happened that by order of these lords the couriers started on their
journeys at the same time. It is required to find in how many days
they will meet.'
The Nuns in their Cells
Pacioli in his De Viribus (c. 1500) posed this problem of rearrange-
ments:
30 Penguin Book of Curious and Interesting Puzzles
95. There are eight nuns, one in each cell, making a total of three
nuns along each side of the courtyard. How can they be rearranged
so that there are four nuns along each side?
1 1 1
1 1
1 1 1
•
Nicolas Chuquet was a doctor by profession, and also the best French
mathematician of his time. These two problems are from his Triparty
en La science des nombres, published in 1484.
96. A carpenter agrees to work on the condition that he is paid £2 for
every day that he works, while he forfeits £3 every day that he does
not work. At the end of thirty days he finds he has paid out exactly as
much as he received. How many days did he work?
97. Liquid Pouring This problem first appears in Chuquet. You
have two jars holding 5 and 3 pints respectively, neither jar being
marked in any way. How can you measure exactly 4 pints from a
cask, given that you are allowed to pour liquid back into the cask?
The Josephus Problem
Chuquet was also the first to present an incident In the life of
Josephus as a problem:
98. Josephus, during the sack of the city of Jotapata by the Emperor
Vespasian, hid in a cellar with forty other Jews who were determined
to commit suicide rather than fall into the hands of the Romans. Not
wishing to abandon life, he proposed that they form a circle and that
every third person, counting round the circle, should die, in the order
The Puzzles 31
in which they were selected. In other words, the count was: 'One,
two, three out, four, five, six out . . .' Where did he place himself, and
a companion who also wished to live, in order to ensure that they
were the last two remaining?
When Roman troops were judged to have shown cowardice, they
were lined up and every tenth man picked out for summary execution,
whence our expression 'to decimate' (which, however, has come to
possess the stronger meaning of 'reducing to one tenth' of the original
number).
This snatch of military history may be the source for the Josephus
problem, based on an incident first described by the unknown author
of the early work De Bello ludaico. The author describes how
Josephus, the historian of the Jewish struggle against the Romans,
once saved himself by just this trick.
Later versions pitted Christians against their enemy of the period,
the Turks:
99. On board a ship, tossed by storms and in danger of shipwreck,
are fifteen Christians and fifteen Turks. To lighten the load and save
the ship, half are to be thrown overboard. One of the Christians
suggests that all should stand in a circle and every ninth person
counting round the circle should be chosen. How should the Chris-
tians arrange themselves in the circle to ensure that only the Turks
die?
100. In a later Japanese version, thirty children, sons of the same
father by his first and second marriages, are too numerous to share
his inheritance. So the second wife suggests that the children be
placed in a circle and eliminated by counting continually around the
circle, eliminating every nth child.
By malice, she ensures that the first fourteen children to be elimin-
ated are all the sons of the first wife. The remaining child of that
wife, seeing that he is alone, suggests that the order of counting now
be reversed, and the second wife agrees, confident that one of her
children must be the last survivor, but to her mortification, all her
own children are then eliminated.
How were the children arranged, and how was the count done?
32 Penguin Book of Curious and Interesting Puzzles
Exchanging the Knights
This is one of the earliest recreative chess problems, posed by Guarini
di Forli in 1512.
~ ~
~ ~
101. Two white knights and two black knights are placed at the
opposite corners of this portion of a chessboard. How can the white
knights take the places of the black knights, and vice versa, moving
according to the rules of chess?
•
Niccoli> Fontana (c. 1499-1557), nicknamed Tartaglia (the Stammerer),
was the brilliant mathematician who discovered how to solve the
cubic equation, only to have Cardano wheedle the solution out of
him and publish it himself.
These problems are from his General Trattato of 1556 and Quesiti
et Inventioni Diverse of 1546.
102. A man has three pheasants that he wishes to give to two fathers
and two sons, giving each one pheasant. How can it be done?
103. A man dies, leaving seventeen horses to be divided among his
heirs, in the proportions i: i : ~ . How can this be done?
Tartaglia also gives a problem of this type:
104. A dishonest servant removes 3 pints of wine from a barrel,
replacing them with water. He repeats his theft twice, removing in
total 9 pints and replacing them with water. As a result the wine
The Puzzles 33
remaining in the barrel is of half its former strength. How much wine
did the barrel originally hold?
Bachet
Claude Gaspar Bachet de Meziriac (1581-1638) was a poet and
translator, and one of the earliest members of the Academie Fran\;aise
as well as a mathematician.
He is famous for two works, his Greek text of the Arithmetica of
Diophantos (1621), accompanied by his own Latin commentary, and
the first European work devoted to mathematical recreations, Prob-
lemes plaisans et delectables qui se font par les nombres (1612).
The Problemes plaisans was largely a compendium of previous
puzzles. It contains river-crossing problems originating with Alcuin, a
method of constructing magic squares which is that found in Mos-
chopoulos, the Josephus problem as solved by Tartaglia, a liquid-
pouring problem, and several think-of-a-number tricks, which are
presented here in the form of problems: how is the original number
recovered after the following operations?
105. A person chooses secretly a number, and trebles it, telling you
whether the product is odd or even. If it is even, he takes half of it, or
if it is odd, he adds one and then takes one half. Next he multiplies
the result by 3, and tells you how many times 9 will divide into the
answer, ignoring any remainder. The number he chose is - what?
106. The subject chooses a number less than 60 and tells you the
remainders when it is divided by 3, 4 and 5, separately, not succes-
sively. The original number is - what?
107. The first person secretly chooses a number of counters, greater
than 5, and the second person takes three times as many. The first
then gives 5 counters to the second, who then gives the first three
times as many counters as the first has in his hand. You can now say
that the second person has - how many - counters in his hand?
Problemes plaisans also contains the famous problem of the weights.
Weights for use with a balance were traditionally made in nested
form, so that one weight fitted inside the other and the largest weight
contained all the smaller weights. Modern sets of weights in which
each fits snugly into the top of the next in the series are a variation of
34 Penguin Book of Curious and Interesting Puzzles
this. It was natural to wonder how many weights, and which weights,
were really necessary to weigh a given quantity. Bachet asked:
108. What is the least number of weights that can be used on a scale
pan to weigh any integral number of pounds from 1 to 40 inclusive, if
the weights can be placed in either of the scale pans?
Bachet's Diophantos is most famous because it was in the margin of
his own copy that Pierre de Fermat wrote a comment on Diophantos's
Book II, problem 8, to solve xr + yZ = a
Z
in integers: 'it is impossible
to separate a cube into two cubes, or a biquadrate [fourth powers)
into two biquadrates, or in general any power higher than the second
into two powers of like degree; I have discovered a truly remarkable
proof which this margin is too small to contain.'
This theorem became known as Fermat's Last Theorem, and
remains unresolved to this day, though it is widely suspected that it is
true.
In his commentary on Diophantos VI, 18, Bachet asked for a
triangle with rational sides and a rational altitude, which means that
the triangle also has a rational area. Because the area of a triangle can
be calculated from the sides using Heron's formula,
A = Js(s - a)(s - b)(s - c)
where s = !(a + b + c), such triangles are called 'Heronian'. Since
all the measurements can be multiplied up to make them integers,
Heronian triangles are often considered to have integral sides and
area.
109. What are the sides and area of the unique Heronian triangle,
one of whose altitudes and its three sides are consecutive numbers?
110. What are the three Heronian triangles, which are not right-
angled, whose area and perimeter are equal?
111. The area of a Heronian triangle is always a multiple of 6. What
is the unique Heronian triangle with area 24?
Henry van Etten
Henry van Etten (1624) was the author of Mathematical Recreations,
Or a Collection of sundrie excellent Problemes out of ancient and
The Puzzles 35
moderne Phylosophers Both usefull and Recreative, published in
French in 1624 and first published in English translation in 1633.
It was a compilation, naturally, including questions from the Greek
Anthology and copying from Bachet, on which it was certainly based,
but containing much extra and varied material.
It is also an important early work on conjuring, containing the first
description of the 'Inexhaustible Barrel' or 'Any Drink Called For',
which allows a variety of drinks to be poured at the magician's whim
from the same spout.
Its mathematical problems are mixed up with mechanical puzzles
and experiments in optics and hydrostatics, instructions on the making
of fireworks, and tips such as 'How to keep wine fresh without ice or
snow in the height of summer'.
The first mechanical problem is to break a staff resting on two
glasses of water, attributed to Aristotle. The solution is to hit it
sufficiently sharply in the middle, and it will break, due to the inertia
of the staff.
112. Arrange three knives so that they 'hang in the air without being
supported by anything but themselves'.
Variants 111 Victorian puzzle books demanded how three knives might
be used to support a drinking glass, in the ample space between three
other drinking glasses placed on the table with more than enough
space for a fourth glass to be placed on the table between them.
Several of the following problems also appear two centuries later
as popular Victorian amusements.
113. How can a stick be made to balance securely on the tip of a finger?
114. You have a strong staff, and a bucket almost full of water.
Required to support the bucket over the edge of the table.
115. How can a bottle be lifted using only a single straw?
116. What shape of bung can be used to plug three different holes,
one square, one triangular and one circular?
117. How maya man have his head upwards and his feet upwards at
the same time?
36 Penguin Book of Curious and Interesting Puzzles
118. Two men ascend two ladders, at the same speed, and yet they
get further apart. Explain.
119. Where can a man look south in all directions?
120. How can a compass with a fixed opening be used to draw
circles of different sizes?
121. How can an oval be drawn with one turn of the compass?
122. Two horses were born at the same time, travelled the world,
and then died at the same time, but did not live to the same age. How
was this possible?
123. 'Three women, A, B, C, carried apples to a market to sell. A had
20, B, 30, and C, 40; they sold at the same price, the one as the other;
and, each having sold all their apples, brought home as much money
as each other. How could this be?'
124. Why must there certainly be at least two people in the world
with exactly the same number of hairs on their head?
•
Pierre de Fermat (1601--65) was a lawyer by profession and an
amateur mathematician of genius who contributed to the development
of the calculus and the invention of analytical geometry, and who
leapt beyond Diophantos to found the modern theory of numbers.
He posed the following problem to Torricelli, Galileo's famous
pupil, who invented the barometer:
125. Find the point whose sum of distances from the vertices of a
given triangle is a minimum.
This problem has a natural appeal, because it can be interpreted as
asking for the shortest road network that will join three towns at the
vertices of the triangle. The next problem occurs first in Urbino
d' Aviso's treatise on the sphere (1682):
126. A strip of paper can be transformed into a pentagon. How?
Prince Rupert's Cube Prince Rupert was a nephew of Charles I of
England, a soldier in the Civil War and an inventor and early member
of the Royal Society. He enquired:
The Puzzles 37
127. What IS the largest cube that can be passed through a square
hole cut in a given cube?
Sir Isaac Newton (1642=1727) composed a book on elementary alge-
bra, his Arithmetica Universalis (1707), in which this problem occurs:
128. If a cows graze b fields bare in c days,
and a' cows graze b' fields bare in c' days,
and a" cows graze b" fields bare in c" days,
what is the relationship between the nine magnitudes a to c"?
In 1693 Samuel Pepys the diarist and Secretary for the Navy wrote to
Newton with this query, a natural question for a gambler:
129. Which is more likely, to throw at least 1 six with 6 dice, at least
2 sixes with 12 dice, or at least 3 sixes with 18 dice?
The misaddressed letters
Niclaus Bernoulli (1687-1759), one of the extraordinary Bernoulli
family which produced nine outstanding mathematicians in three
generations, considers this problem (but with n letters instead of ten):
130. A correspondent writes ten letters and addresses ten envelopes,
one for each letter. In how many ways can all the letters be placed in
the wrong envelopes?
131. A related question: if seven letters are placed in seven envelopes
randomly, how many letters would you expect, on average, to find in
their correct envelopes?
•
Leonhard Euler (1707-83) was one of the most versatile mathemati-
cians of all time, as well as one of the greatest. Here are three of the
problems he considered.
132. The Knight's Tour How can a knight make a complete tour of
the chessboard shown on p. 38, visiting each square once and only
once, and ending up a knight's move from its starting square - so that
the circuit is continuous?
38 Penguin Book of Curious and Interesting Puzzles
The Bridges of Konigsberg In the town of Konigsberg there were
seven bridges across the river Pregel. This popular question was
answered by Euler in 1736:
133. Is it possible to go for a walk, crossing each bridge once, bur not
crossing any bndge twice?
This was the first ever problem in what is now called graph theory. A
graph is a set of points, called vertices or nodes, joined by a set of
lines, called edges. A vertex where an odd number of edges meet is
called an odd vertex, naturally. Graph theory poses many problems,
some of them very simple and simply puzzling:
The Puzzles 39
134. Why is the number of odd vertices in a graph always even?
The Thirty-six Officers Problem Euler considered the problem of
placmg thirty-six officers, comprismg a colonel, lieutenant-colonel,
major, captain, lieutenant and sub-lieutenant from each of six regi-
ments, in a square array so that no rank or regiment will be repeated
in any row or column.
This problem turns out to be impossible, but the same problem
.with twenty-five officers IS not:
135. How can five each of As, Bs, Cs, Ds and Es be placed in these
cells so that no letter is repeated in any row or column?
The Ladies' Diary or Woman's Almanac, 1704-1841
The Ladies' Diary was first published in 1704 and consisted initially
of recipes, sketches of notable women, and articles on education and
health, naturally appealing to its readership.
Within a short time, however, its contents changed, to be replaced
by rebuses, enigmas and mathematical questions. That it not only
survived but flourished is a blow in the eye to those who suppose that
women cannot be interested in mathematics, and proof that carica-
tures of women as mathematically incapable were less well-established
in the early eighteenth century than the late twentieth century. Al-
though men soon proposed and answered many of the questions,
40 Penguin Book of Curious and Interesting Puzzles
women continued to contribute as posers and solvers.
The problems were initially proposed, in the manner of the times,
in verse, but, mathematics not lending itself to versification, this
practice was soon abandoned.
Although it was not a compilation, some venerable problems were
proposed. The first question that is identified as 'Solution by a Lady'
(respondents were often anonymous or identified by aliases, such as
'Anne Philomathes') concerned grains of wheat on a chessboard, as
payment for sixty-four diamonds: one grain on the first square, two
on the second and so on.
In subsequent years, many of the questions were very difficult, and
were answered by almost all the famous English mathematicians of
the eighteenth century.
The Mathematical Questions Proposed in the Ladies' DIary, and
their Original Answers, together with Some New Solutions, from its
Commencement in the Year 1704 to 1816 was a compilation, the
work of Thomas Leybourn, a professor at the Royal Military Col-
lege. The very first mathematical question, posed in the year 1707,
was:
136. 'In how long a time would a million of millions of money be in
counting, supposing one hundred pounds to be counted every minute
without intermission, and the year to consist of 365 days, 5 hours, 45
minutes?'
This early question in verse illustrates the difficulties of the form:
137. If to my age there added be,
One half, one third, and three times three;
Six score and ten the sum you'll see,
Pray find out what my age may be.
138. 'A person remarked that upon his wedding day the proportion
of his own age to that of his bride was as 3 to 1; but fifteen years
afterwards the proportion of their ages was 2 to 1. What were their
ages upon the day of their marriage?'
Question 36 was posed by Mrs Barbara Sidway:
139. 'From a given cone to cut the greatest cylinder possible.'
Question 42 concerns a maypole which breaks, the tip making a mark
on the ground. In other words, it is a variant of problem 66 and more
The Puzzles 41
than 1000 years old. Question 51 was also old. The solution noted
that the problem appeared in Diophantos, Book V.
In contrast the next three problems have a modern feel:
140. What is the least number which will divide by the nine digits
without leaving a remainder?
141. 'There came three Dutchmen of my acquaintance to see me,
being lately married; they brought their wives with them. The men's
names were Hendrick, Claas, and Cornelius; the women's Geertrick,
Catriin, and Anna; but I forget the name of each man's wife.
'They told me that they had been at market, to buy hogs; each
person bought as many hogs as they gave shillings for each hog;
Hendrick bought twenty-three hogs more than Catriin, and Claas
bought eleven more than Geertrick; likewise, each man laid out 3
guineas more than his wife. I desire to know the name of each man's
wife?' (A guinea was 21 shillings.)
142. 'Being at so large a distance from the dial-plate of a great clock,
that I could not distinguish the figures; but as the hour and minute
hands were very bright and glaring,' the correspondent noted that
they were in a straight line and pointing upwards to the right. It was
evening. What was the time?
•
The Vanishing Square Paradox
William Hooper, in his Rational Recreations (1774), proposed the first
of many vanishing square paradoxes.
8 5
5 8
3 3
8
42 Penguin Book of Curious and Interesting Puzzles
143. The top square has an area 8' = 64. The same four pieces,
when reassembled to make the lower figure, form a rectangle 5 x
13 = 65. Where has the extra square come from?
144. This is a modern variant. Your task is to reassemble these
sixteen pieces to make another 13 x 13 square, but with an empty
square in the centre.
Rowing with and against the Tide
This is another first, which occurs in an arithmetic textbook published
in the United States in 1788.
145. 'If, during ebb tide, a wherry should set out from Haverhill, to
come down the river, and, at the same time, another should set out
from Newburyport, to go up the river, allowing the difference to be
18 miles; suppose the current forwards one and retards the other 1t
miles per hour; the boats are equally laden, the rowers equally good,
and, in the common way of working in still water, would proceed at
the rate of 4 miles per hour; when, in the river, will the two boats
meet?'
Rational Amusements for Winter Evenings
John Jackson was 'A Private Teacher of Mathematics' who decided
that there were many puzzles scattered around, but not collected
The Puzzles 43
together in one small and convenient volume, so he assembled them
himself and wrote Rational Amusements for Winter Evenings, or, A
Collection of above 200 Curious and Interesting Puzzles and Para-
doxes relating to Arithmetic, Geometry, Geography, &, 'Designed
Chiefly for Young Persons', which appeared in London in 1821.
From its great'rarity it may be inferred that it did not sell many
copies, which is a shame because, in addition to its superb title -
anticipating the present volume! - it contains many of the classic
puzzles and some that had not apparently appeared earlier in print, in
particular a collection of ten tree-planting problems and a collection
of fifteen variously shaped tiles which resemble a complicated set of
Tangram pieces, to be assembled to form a square, a rectangle, a
right-angled triangle, a rhombus, and so on.
146. 'It is required to express tOO by four 9s.'
147. 'If from six ye take nine, and from nine ye take ten
(Ye youths, now the mystery explain),
And if fifty from forty be taken, there then,
Shall just half a dozen remain.'
148. 'Place the nine digits, so that the sum of the odd digits may be
equal to the sum of the even ones.'
149. 'One third of twelve if you divide,
By just one fifth of seven,
The true result (it has been tried)
Exactly is eleven.'
150. 'Place in a row nine (digits] each different from the others.
Multiply them by 8, and the product shall still consist of nine
different (digits].'
151. You have 12 pints of wine in a barrel and you wish to divide it
into 6 pints for a friend and 6 pints for yourself, but you only have
containers holding 7 and 5 pints. How can you succeed?
152. 'With the first nine terms of the geometrical progression 1, 2, 4
... to form a product of 4096 each way.' (In other words, the
products of the numbers of each vertical column and each horizontal
row in the box overleaf must equal 4096.)
44 Penguin Book of Curious and Interesting Puzzles
153. 'With the numbers 1,2,3, ... , to 16, to form 34 every way.'
154. 'A Cheshire cheese being put into one of the scales of a false
balance, was found to weigh 16 lbs, and when put into the other only
9 lbs. What is the true weight?'
155. 'Mathematicians affirm that of all bodies contained under the
same superficies, a sphere is the most capacious; but they have never
considered the amazing capaciousness of a body, the name of which
is now required, of which it may be truly affirmed that supposing its
greatest length 9 inches, greatest breadth 4 inches, and greatest depth
3 inches, yet under these dimensions it contains a solid foot?'
The Puzzles 45
156. 'Divide a circle into four equal parts by three lines of equal
length.'
157. 'To make a triangle that shall have three right-angles.'
158. 'To inscribe a square in a given circle, by means of compasses
only, supposing the centre to be known.'
The next question asks the reader to inscribe a regular dodecagon
(12 sides) in a circle under the same conditions.
Jackson'S book concludes with no less than sixty 'Geographical
Paradoxes', certainly well-calculated to enliven lessons on the globe.
One of these depends on acquaintance with a specific physical phenom-
enon, and is worth relating to illustrate how puzzles in natural
philosophy were often mixed in with more logical or mathematical
puzzles:
'There is a certain place in Great Britain, where, when the
tide is in, the sheep may be seen feeding on a certain
neighbouring island; yet, when the tide is out, and the water at
the lowest, not one can be seen, though they be feeding there at
the same instant.'
jackson's explanation is: 'The place may be the wharf at Green-
wich, the Isle of Dogs over against it, and the appearance caused by
refraction, when the water is high.'
The remaining puzzles require no such localized knowledge.
159. 'There are three remarkable places on the globe that differ in
latitude, as well as in longitude; and yet, all of them lie under the
same meridian.'
160. 'There is a particular place on earth, where the winds (although
frequently veering round the compass) always blow from the north
point.'
161. 'There is a certain village in the Kingdom of Naples, situated in
a very low valley, and yet the sun is nearer to the inhabitants thereof,
every noon by 3000 miles and upwards, than when he either rises or
sets, to those of the said village.'
162. 'There is a certain island, situated between England and France,
and yet, that island is farther from France than England is.'
46 Penguin Book of Curious and Interesting Puzzles
163. Christians the week's first day for sabbath hold,
The Jews the seventh, as they did of old,
The Turks the sixth, as we have oft been told.
How can these three, in the same place, and day,
Have each his own true sabbath, tell, I pray.
164. A traveller sets out on a journey, and eventually returns to the
place from where he started. During his journey, his head has travelled
12 yards further than his feet, and yet his head remains attached to
his body. How is this possible?
From Ozanam to Hutton
Jacques Ozanam was a Frenchman, born in 1640, who wrote a book
of recreations based on Bachet and other traditional sources. It was
greatly enlarged and improved by Jean Etienne Montucla (born
1725), a friend of Diderot and D' Alembert.
It was translated into English by Charles Hutton, professor at the
Royal Military Academy at Woolwich. Edward Riddle's revision of
it, called Recreations in Mathematics and Natural Philosophy (1840),
was the largest collection of mathematical puzzles and recreations
published in this country up to that time. The following problems are
taken partly from Ozanam's 1741 edition, and partly from Riddle.
165. The hour and minute hands of a watch coincide at noon. When
will they once again coincide, during the next 12 hours?
166. 'We are told by Father Sebastian Truchet, of the Royal Academy
of Sciences, in a memoir printed [in) 1704 ... that having seen during
the course of a tour which he made to the town of Orleans, some
square porcelain tiles, divided by a diagonal into two triangles of
different colours ... he was induced to try in how many different
ways they could be joined side by side, in order to form different
figures. [Such tiles) form the object of a pastime, called by the French
Jeu de Parquet ... a small table, having a border round it, and
capable of receiving sixty-four or a hundred small squares ... with
which people amuse themselves in endeavouring to form agreeable
combinations.
How many figures can be formed by three squares if the colours of
the two halves are black and white and if an edge is placed against a
complete edge?
The Puzzles 47
167. If A and B together can complete a task of work in 8 days; and
if A and C together take 9 days, and Band C together take 10 days,
how much will each man take to do the work by himself?
The next problem is equivalent to the dissection of a Greek Cross
into a square:
168. How can five equal squares be dissected and reassembled to
form one large square?
169. Demonstrate Pythagoras's Theorem by dissecting the smaller
squares to form the larger square, the pieces to be moved by transposi-
tion only, without rotation or turning over.
170. DIssect a gIven rectangle into a square.
171. 'A gentleman wishes to have a silver vessel of a cylindrical form,
open at the top, capable of contaIning a cubic foot of liquor; but
being desirous to save the material as much as possible, requests to
know the proper dimensions of the vessel.'
172. A man has two wines, one of which sells at 10 shillings per
bottle, and the other at 5 shillings. What is the mixture that would
sell at 8 shillings a bottle?
173. What is the largest rectangle that can be cut in one piece from
this triangular piece of timber?
The next problem is attributed to a Mr D-, who said that he got it
from M. BuHon, the French naturalist and translator of Newton's
PrincIpia:
48 Penguin Book of Curious and Interesting Puzzles
174. Given any irregular polygon, the mid-points of the sides are
joined in sequence, and this process is then repeated, again and again.
'It is required to find the point where these divisions will terminate.'
175. 'Given two lines and a point within the angle formed by them,
to find the smallest triangle by area that can be cut off.'
176. The Harmonic Square The ancient Greeks considered three
important means, the arithmetic and the geometric, which are well
known, and the harmonic. The harmonic mean of two numbers is
found by taking their reciprocals, finding their average and taking the
reciprocal of the result. In other words the harmonic mean of x and y
is
which simplifies to
1
2xy
x+y
With that explanation, how can the cells of this square be filled so
that the cells in the middle of each side and the centre cell are each
the harmonic means of the numbers sandwiching them? The central
number is sandwiched, of course, in four different ways.
177. One player chooses a number less than 11. The second player
does likewise and adds his number to the first player's number. The
The Puzzles 49
first player again adds a number less than 11, and so on. The player
who reaches the grand total of 100 or more is the winner. Is there a
winning strategy?
178. The Eight Queens How can eight queens be placed on a
chessboard so that no queen attacks any other?
This problem was first posed by Max Bezze!, wntlng under the
pseudonym 'Schachfreund', in the chess magazine Berliner Schachzei-
tung, in 1848. To find all the solutions is extremely difficult, because
of the size of the board. An easier problem is:
179. How can 4 (5,6) queens be placed on a 4 x 4 (5 x 5, 6 x 6)
board so that no queen attacks any other?
The Victorian Era
Between John Jackson, Riddle's edition of Ozanam, and the end of
the century, a wealth of books appeared, with titles such as The
Games Book for Boys and Girls, Cassell's Book of Indoor Amuse-
ments, Card Games and Fireside Fun, and The Illustrated Book of
Puzzles and Parlour Pastimes: A Repertoire of Acting Charades, Fire-
Side Games, Enigmas, Riddles, Charades, Conundrums, Arithmetical
and Mechanical Puzzles, Parlour Magic etc.
They were mostly written for young people, and contained sections
of mathematical, mechanical and word puzzles, often a section of
magic tricks, simple scientific experiments, plus a wealth of literary
puzzles, enigmas, charades, rebuses and maybe chapters on outdoor
as well as 111door games and amusements.
The authors showed the usual reliance on old sources, which IS
why the problems below are not credited to particular books.
According to Dudeney, writing in his The World's Best Puzzles, this
puzzle has been attributed to Sir Isaac Newton, but Dudeney himself
knew of no earlier source than 'a rare book, published in 1821',
which was John Jackson's P.ational Amusements:
180. Tree in a Row How can nine trees be arranged in ten rows
with three trees in every row?
Parlour Pastimes posed a variant in verse, with an extra condition:
50 Penguin Book of Curious and Interesting Puzzles
181. Ingenious artist, pray dispose,
Twenty-four trees in twenty-eight rows;
Three trees I'd have in every row,
A pond in the midst I'd have also;
A plan of it I fain would have,
Which makes me your assistance crave.
The following, on the face of it, is a simpler problem:
182. 'Plant four trees at equal distance from each other.'
183. 'Place twelve counters in six rows in such a manner that there
shall be four counters in each row.'
184. You have to divide the number 45 into four parts. To the first
part you add 2, from the second part you take 2, the third part you
multiply by 2, and the fourth part you divide by 2, so that the sum of
the addition, the remainder of the subtraction, the product of the
multiplication, and the quotient of the division are all equally and
precisely the same. How is this possible?
185. 'Having placed eight coins in a row, as under, show how they
can be laid or placed in four couples, removing only one at a time,
passing over two each time.'
2 3 4 5 678
186. 'Draw six lines as under, add five other lines, and make the
whole form nine.'
(There is also a French version of this puzzle: add three lines to make
eight.)
187. The half of twelve IS seven, as I can show;
The half of thirteen eight; can thiS be so?
188. These dogs are dead, perhaps you'll say;
Add four lines, and then they'll run away.
The Puzzles 51
189. 'Of five pieces of wood, or paper, cut in the following shapes,
form a cross.'
52 Penguin Book of Curious and Interesting Puzzles
190. From these five shapes, also form a cross:
191. 'A charitable individual built a house in one corner of a square
plot of ground, and let it to four persons. In the ground were four
cherry trees, and it was necessary so to divide it, that each person
might have a tree and an equal portion of garden ground. Here is a
sketch of the plot. How is it to be divided?'
The next four problems are from Scientific Amusements by 'Tom
Tit', based on the popular French book La Science Amusante.
192. How can an equilateral triangle be constructed by folding a
square of paper?
193. How can a ladder be made out of a single sheet of paper,
without lIsing gum or other adherent, and this to be effected with
three cuts with the scissors?
The Puzzles 53
194. How can a half-crown be passed through a hole the size of a
shilling? The old half-crown was approximately 3.1 cm across and
the old shilling was approximately 2.3 cm in diameter.
195. What is the largest envelope that can be constructed by folding a
rectangle of paper?
196 .• A carpenter had to mend a hole in the floor which was two feet
wide and twelve feet long. The board given him to mend it was three
feet wide and eight feet long.'
How can he achieve this feat, cutting the board into only two
pieces?
197. How can this board, marked as shown, be cut into four identical
pieces, so that each piece contains three of the marks, and no mark is cut?
o o o
o
00
o
00
o o o
198. Cut a hole in a visiting card large enough for a person to climb
through.
199. Place ten coins in a row upon a table. Then, taking up anyone
of the series, place it upon some other, with this proviso, that you
pass over just two coins. Repeat this till there is no single coin left.
200. Arrange the digits 0 to 9 so that they sum to 100.
54 Penguin Book of Curious and Interesting Puzzles
201. How many animals are concealed in this picture?
202. This is another square with one quarter missing.
How can it be divided between four sons, so that each receives an
area identical in shape and size to the others'?
203. How many strokes are necessary to draw this figure, without
going over any line twice? A stroke ends as soon as you lift your
pencil from the paper.
The Puzzles 55
204. A tumbler is resting on three lOp pieces on a table cloth. Under
the centre of the tumbler lies a 20p coin. Being thinner than the lOp
piece, it can in theory be removed from under the edge of the tumbler
without disturbing the tumbler, but you are not allowed to use a
knife, or sheet of paper or card or any other suitably thin instrument.
How do you remove the 20p piece?
205. Here are three squares, each composed of four matches. Make
them into one by taking one match away, and moving only three
others.
DDD
206. Here are the same three squares of matches.
DDD
Move three matches to show what matches are made of.
207. Taking one corner of a plain unknotted handkerchief in one
hand, and the opposite corner in the other hand, you bring the
corners together, and then apart, and Lo and Behold! there is a knot
in the handkerchief!
At no time did you release either of the corners of the handkerchief,
56 Penguin Book of Curious and Interesting Puzzles
which remain between the fingers of each hand, exactly as you picked
it up. How is this possible?
208. Here is a correct addition sum. Your puzzle is to cover one of
the numbers completely, to leave a new addition sum which still
totals 1240.
3/8
303
300
lOY-
2..IS
12'10
209. An Easy Solitaire Each number represents a piece that can
jump over any other piece, either vertically, horizontally or diagonally,
into an empty square beyond. How can all the pieces be removed,
except one, which shall be the 9, which ends up in its original
position in the centre?
1 8 7
2 9 6
3 4 5
The Puzzles 57
210. How can four triangles be made with just six matches?
211. These twelve counters are arranged to form six equal squares.
Remove just three counters to leave just three equal squares.
0 0 0 0
0 0 0 0
0 0 0 0
212. Upon a piece of paper draw
The three designs below;
I should have said of each shape four,
Which when cut out will show,
If joined correctly, that which you
Are striving to unfold -
An octagon, familiar to
My friends both young and old.
3
58 Penguin Book of Curious and Interesting Puzzles
The most comprehensive compilation of this period was Everybody's
Illustrated Book of Puzzles, selected by Don Lemon (1890), which
crammed 794 puzzles, a large majority of them word puzzles, into 125
pages and included most of the puzzles above, and the following. The
expression of puzzles in verse was typically Victorian, as was the
delight in quibbles and trick questions.
213. There was a poor man called johannes Bull,
Who children did possess, a quiver full;
And who yet managed somehow to scratch on,
By the true help of daughter and of son.
Six little workers had he, each of whom
Earned something for the household at the loom.
I will not tell you how much each did gain,
For I'm a puzzler, and I don't speak plain;
But, as I would you should possess a clew,
Some tell-tale facts I'll now disclose to you.
Week after week, jane, Ann, joe, Bet, Rose, jim,
Earn ten and tenpence, father says, for him,
And in this way: The eldest daughter, jane,
Gains seven pence more than sister Anne can gain;
Ann eightpence more than joe; while joe can get
By his endeavours sixpence more than Bet;
Bet, not so old, earns not so much as those,
But by her hands gets fourpence more than Rose;
Rose, though not up to jane, yet means to thrive,
And every week beats jim by pennies five.
Now, say what each child worker should receive
When father draws the cash on pay day eve?
214. Who Can Tell?
Twice six are eight of us,
Six are but three of us,
Nine are but four of us,
What can we possibly be?
Would you know more of us?
I'll tell you more of us.
Twelve are but six of us,
Five are but four of us, now do you see?
The Puzzles 59
215. A row of four figures in value will be
Above seven thousand nine hundred and three;
But when they are halved, you'll find very fair
The sum will be nothing, in truth I declare.
216. Quibbles
(a) Add the figure 2 to 191 and make the answer less than 20.
(b) How can I stretch my hands apart, having a coin in each hand,
and without bringing my hands together, cause both coins to come
into the same hand?
(c) How must I draw a circle round a person placed in the centre of a
room so that he will not be able to jump out of it, though his legs
should be free?
(d) If five times four are thirty-three, what will the fourth of twenty
be?
217. A box has nine ears of corn in it. A squirrel carries out three
ears a day, and yet it takes him nine days to carry the corn out. How
is this explained?
218. 'A person let his house to several inmates and, having a garden
attached to the house, he wished to divide it among them. There were
ten trees in the garden and he desired to divide it so that each of the
five inmates should have an equal share of the garden and trees. How
did he do it?'
! !
60 Penguin Book of Curious and Interesting Puzzles
219. This is a trick for a teenager addressing an adult. Let the
teenager subtract his or her age from 99, then ask the older person to
add this difference to their own age, and then to take the first digit of
the amount and add it to the remaining figure.
Query: what will the answer tell the younger person?
The Learned Professor Hoffman
Professor Hoffman's real name was the Reverend Angelo John Lewis.
His most famous book, Puzzles Old and New (1893), was chiefly
devoted to the many popular mechanical puzzles but he also included
other Victorian favourites. He also wrote on magic and conjuring.
220. How can this rectangle with two tabs be cut into two pieces to
make a complete rectangle?
221. 'Required, of the numbers, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, to compose
two fractions, whose sum shall be equal to unity. Each number to be
used once, and once only.'
222. 'Required, to find a number of six digits of such a nature that if
you transfer the two left-hand digits to the opposite end, the new
number thus formed is exactly double the original number.'
223. 'A man goes into a shop and buys a hat, price one guinea. He
offers in payment for it a £5 note. The hatter gets the note cashed by
a neighbour, the purchaser pocketing his change, £3 19s, and walking
off with the hat. No sooner had he left, however, than the neighbour
comes in with the news that it is counterfeit, and the hatter has to
refund the value.'
The Puzzles 61
How much is the hatter out of pocket by the transaction? (A
guinea was 21 shillings.)
224. 'Fifteen matches being laid on the table so as to form five equal
squares, required, to remove three matches so as to leave three such
squares only.'
225. How can three matches be taken away to leave a total of seven
triangles behind?
/\
~
LV\L\
226. An old gentleman was asked who dined with him on Christmas
day.
'Well, we were qUite a family party,' he replied; 'there was my
father's brother-in-law, my brother's father-in-law, my father-in-Iaw's
brother-in-law, and my brother-m-Iaw's father-in-law.'
It afterwards transpired that he had dined alone, and yet his
statement was correct.
How could this be?
62 Penguin Book of Curious and Interesting Puzzles
227. What is the difference between six dozen dozen and a half a
dozen dozen?
228. Five herrings were divided between five persons. Each had a
herring, and yet one remained in the dish. How was this possible?
Mathematicians have often created problems of a popular nature in
between their more 'serious' work. Euler was an example. The
followl11g examples come from three famous nineteenth-century math-
ematicians, and one anonymous examination paper.
Hamilton and the Icosian Game
The Icosian Game was invented by W. R. Hamilton, the famous
mathematician, and sold to J. Jacques and Son, makers of fine chess
sets, for £25. It was published in London in 1859.
229. As 'The Traveller's Dodecahedron' it consisted of a regular
dodecahedron, handsomely made in wood, with the names of twenty
cities marked at the vertices, in alphabetical order from B for Bruxelles
to Z for Zanzibar, with a few letters omitted. They were jOl11ed by
black lines along the edges, indicating the routes between them, and
the object was to visit every town once and only oncc. How can this
be done?
A solid model is not necessary: the arrangement of the towns is
indicated 111 this figure, under which form the puzzle was known as
the Icoslan Game.
R
w B
The Puzzles 63
The First Pursuit Problem The first pursuit problem appeared in a
Cambridge Tripos paper of 5 January 1871. It concerned three bugs,
each chasmg the next. Here is a variation:
230. Four dogs start from the four corners of a square field, of side
100 yards, and race towards each other with constant speed of 3
yards per second, all the dogs starting off in a clockwise direction.
Where will they meet, and how long will it take them to catch each
other?
231. The Age of Augustus de Morgan 'Writing in 1864, Professor de
Morgan said he was x years old in the year Xl AD. When was he born?'
Rouse Ball's Mathematical Recreations and Essays
Rouse Ball was the original author of the famous Mathematical
Recreations and Essays. The first edition was published in 1892, and
contained chapters in arithmetical and geometrical recreations, mech-
anical problems, magic squares and unicursal puzzles, as well as
the Essays of the title on the Cambridge Mathematical Tripos, astrol-
ogy, hyper-space, cryptography and cyphers, and other subjects.
In subsequent editions, the recreational material became predomi-
nant.
232. The figure represents a portion of a chessboard. In the top left
corner are eight white pawns, and in the lower right corner, eight
black pawns. One mbve consists of moving a pawn into the adjacent
a b c
d e f
9
h • H G
F E D
C B A
64 Penguin Book of Curious and Interesting Puzzles
empty square either horizontally or vertically, or jumping a piece over
an adjacent piece, again horizontally or vertically, into the empty
square immediately beyond.
No diagonal or backward moves are permitted. How can the black
and white pawns be exchanged in the minimum number of moves?
Sylvester and the Postage Stamp Problem
J. J. Sylvester (1814-97) sent the following puzzle to the Educational
Times, a journal famous in its day for its mathematical problems and
the eminent mathematicians who contributed:
233. I have a large number of stamps to the values of Sp and 17p
only. What is the largest denomination which I cannot make up with
a combination of these two different values?
The Tower of Hanoi and Other Puzzles
Edouard Lucas (1842-91) was a mathematician who studied the
Fibonacci sequence and the Lucas sequence, which was named in his
honour.
In his Recreations mathematiques Lucas discussed Sam Loyd's
'Fifteen' puzzle under the title Le Jeu du Taquin and then generalized
it to consider any arrangement of squares. Here is a simple case, a
rectangular circuit with two additional squares.
234. Four pieces occupy the shaded squares as shown. Is it possible
to exchange C for D and also A for B?
I J
-
-
C D
- -
I J
A B
The Puzzles 65
Inspired perhaps by his Jeu du Taquin, or merely in a delayed
response to the coming of the railways, Lucas posed the first shunting
problems:
235. Train A requires to overtake train B, making use of the cul-de-
sac line, which is, however, long enough to contain only half of train
B. How would you advise the drivers?
Voie in cul-de-sac
Voie principale
--
A B
236. A garden is surrounded by a square moat of uniform width.
Wishing to cross the moat, to reach the garden, you pick up two
planks, each 8 feet long, but the moat is 10 feet wide. Can you cross
it safely?
237. Every day at noon a ship leaves Le Havre for New York and
another ship leaves New York for Le Havre. The trip lasts seven days
and seven nights. How many New York-Le Havre ships will the ship
leaving Le Havre today meet during its journey to New York?
238. Lucas's best-known invention is his 'Tower of Hanoi', which
was presented to the public in 1883 as the creation of Mr Claus, of
the College of Li-Sou-Stian, anagramming Lucas and the Lycee Saint-
Louis, where he was then teaching. Hanoi was the capital of Vietnam,
an exotic and faraway country that was also a French colony.
66 Penguin Book of Curious and Interesting Puzzles
The year after its publication, the following story was published,
not by Lucas, to explain the puzzle:
In the great temple at Benares, says he, beneath the dome
which marks the centre of the world, rests a brass plate in
which are fixed three diamond needles, each a cubit high and
as thick as the body of a bee. On one of these needles, at the
creation, God placed sixty-four discs of pure gold, the largest
disc resting on the brass plate, and the others getting smaller
and smaller up to the top one. This is the Tower of Bramah.
Day and night unceasingly the priests transfer the discs from
one diamond needle to another according to the fixed and
immutable laws of Bramah, which require that the priest on
duty must not move more than one disc at a time and that he
must place this disc on a needle so that there is no smaller disc
below it. When the sixty-four discs shall have been thus
transferred from the needle on which at the creation God placed
them to one of the other needles, tower, temple, and Brahmins
alike will crumble into dust, and with a thunderclap the world
will vanish.
The puzzle is to say how many moves are needed to transfer all
sixty-four discs.
Lewis Carroll (Reverend Charles Ludwig Dodgson)
(1832-98)
Carroll is world-famous as the author of Alice in Wonderland and
Alice Through the Looking Glass, but he was also a witty composer
of puzzles and entertainments, as well as being, least importantly to
The Puzzles 67
the rest of the world, a lecturer in mathematics at Christ Church,
Oxford.
He planned a series of books, never completed, under the general
title Curiosa Mathematica. The second volume was called Pillow
Problems, and illustrated his great ability to solve problems in his
head. But mathematical and logical problems were only a small part
of his output: he invented the word ladder, in which one word is
transformed into another, one letter at a time, as BLACK into
WHITE, and all his writings were riddled with puns, word-play and
logical phantasy.
239. 'A bag contains one counter, known to be either white or black.
A white counter is put in, the bag shaken, and a counter drawn out,
which proves to be white. What is now the chance of drawing a white
counter?'
240. 'If four equilateral triangles be made the sides of a square
pyramid: find the ratio which its volume has to that of a tetrahedron
made of the triangles.'
241. 'Three points are taken at random on an infinite plane. Find the
chance of their being the vertices of an obtuse-angled triangle.'
242. 'I have two clocks: one doesn't go at all, and the other loses a
minute a day: which would you prefer?'
243. The Chelsea Pensioners 'If 70 per cent have lost an eye, 75 per
cent an ear, 80 per cent an arm, 85 per cent a leg: what percentage at
least must have lost all four?'
244. The Two Omnibuses Omnibuses start from a certain point,
travelling In both directions, every 15 minutes. A traveller, starting on
foot along with one of them, meets one coming towards him in 121
minutes: when will he be overtaken by one?
245. 'Supposing on Tuesday, it is morning in London; in another
hour it would be Tuesday morning at the West of England; if the
whole world were land we might go on tracing, Tuesday morning,
Tuesday morning all the way round, till in twenty-four hours we got
to London again. But we know that at London twenty-fours hours
after Tuesday morning it is Wednesday morning. Where, then, in its
passage round the earth, does the day change its name?'
68 Penguin Book of Curious and Interesting Puzzles
246. 'A rope is supposed to be hung over a wheel fixed to the roof of a
building; at one end of the rope a weight is fixed, which exactly
counterbalances a monkey which is hanging on to the other end. Suppose
that the monkey begins to climb the rope, what will be the result?'
247. 'Put down any number of pounds not more than twelve, any
number of shillings under twenty, and any number of pence under
twelve. Under the pounds put the number of pence, under the
shillings the number of shillings, and under the pence the number of
pounds, thus reversing the line. Subtract. Reverse the line again.
Add.'
Query: what was Carroll's conclusion?
This next puzzle is related to the river-crossing conundrum which
goes back to Alcuin, but the situation has been turned on its side.
248. 'A captive Queen and her son and daughter were shut up in the
top room of a very high tower. Outside their window was a pulley
with a rope round it, and baskets fastened at each end of the rope of
equal weight. They managed to escape with the help of this and a
weight they found in the room, quite safely. It would have been
dangerous for any of them to come down if they weighed more than
15 lbs more than the contents of the lower basket, for they would do
so too quick, and they also managed not to weigh less either. The one
basket coming down would naturally of course draw the other up.'
How did they do it?
The Queen weighed 195 lbs, the daughter 1651bs, the son 901bs,
and the weight 75 lbs.
This problem is described by Viscount Simon in his memoir of Lewis
Carroll:
249. One glass contains 50 spoonfuls of brandy and another glass
contains 50 spoonfuls of water. A spoonful of the brandy is transferred
to the water, and the mixture is stirred. A spoonful of the mixture is
then transferred back to the glass of brandy.
Is there now more brandy in the water, or more water in the
brandy?
250. 'Two travellers spend from 3 o'clock till 9 in walking along a
level road, up a hill, and home again; their pace on the level being 4
miles an hour, up hill 3, and down hill 6. Find distance walked: also
(within half an hour) time of reaching top of hill.'
The Puzzles 69
251. 'A customer bought goods in a shop to the amount of 7s 3d.
The only money he had was a half-sovereign, a florin, and a sixpence:
so he wanted change. The shopman only had a crown, a shilling, and
a penny. But a friend happened to come in, who had a double-florin,
a half-crown, a fourpenny bit, and a threepenny bit.
Could they manage it? (A half-sovereign was 10 shillings or 120 pence;
a florin was 2 shillings or 24 pence; a crown was 5 shillings or 60 pence.)
252. The Tangram, this dissection of a square into seven pieces from
which any number of shapes can be composed, goes back at least as
far as the middle of the eighteenth century in China.
It is no surprise that it appealed to Lewis Carroll. Here are four of
the Alice figures as Tangrams. How are they composed?
70 Penguin Book of Curious and Interesting Puzzles
Loyd and Dudeney
Sam Loyd (1841-1911) and Henry Ernest Dudeney (1857-1930) will
always be bracketed as the two greatest puzzle-composers of all time.
Not only did their lives largely overlap, in an era when puzzles of all
kinds were exceptionally popular and newspapers and magazines
were eager to cater for their readers' enthusiasms, but they even
worked together for a short period, often shared each other's ideas
(or sometimes pinched them - Loyd seems to have done rather more
of the pinching) and competed in presenting ingenious new ideas, or
familiar ideas in new dress.
Sam Loyd
Sam Loyd was born in Philadelphia, but his parents soon moved to
New York, where he attended high school. He considered being an
engineer, but gave up the idea when he started to make money from
his puzzles. Loyd was a prodigy whose chess problems alone made
him famous. He was just fourteen when he started to attend a chess
club with his brothers Thomas and Isaac, of whom Isaac also became
a noted problemist. Sam's first problem was published in the same
year, and by the age of sixteen he was problem editor of Chess
Monthly, co-edited by Paul Morphy. But by his late teens he had
already produced the stunning puzzle of the riderless horses, which
the circus owner and showman P. T. Barnum bought from him and
sold as 'P. T. Barnum's Trick Donkey'. Loyd had taken the old puzzle of
the two dogs (problem 188) and given it a brilliant new twist.
Many years later he produced an even more amazing puzzle, the
'Get off the Earth' paradox. This is how he described the circum-
stances of its creation, in the Strand magazine (January 1908, reprinted
in Sam Loyd and His Chess Problems, p. 113).
Unfortunately, it came out in a bad year and did not achieve the
success of some of the others. It was developed under rather odd
conditions. My son, who thinks I can do anything, said to me
one morning, 'Here's a chance, Pop, for you to earn $250,' and
he threw a newspaper clipping to me across the breakfast table.
It was an offer by Percy Williams of that amount for the best
device for advertising Bergen Beach, which he was about to
open as a pleasure resort. I said I would take a chance at it, and
a few days later I had worked out the Chinaman puzzle. It
The Puzzles 71
r - - - - - - - - - - - - - - - ~ · - - - - - - - - - - - - ,
~ - - - - - - - - - - - - - - - - - - - - - - - - - - . - - - ~
Cut out the three rectangles and rearrange them so
that the two jockeys are riding the two horses.
consisted of two concentric pieces of cardboard, fastened
together so that the smaller inner one, which was circular,
moved slightly backward and forward, on a pivot, producing
the mystery. As you looked at them, there were thirteen
Chinamen plainly pictured. Move the inner card around a little
and only twelve Chinamen remained. You couldn't tell what
had become of the other Chinaman, try as you would. Scientists
tried it without success, and indeed no single absolutely correct
analysis was ever submitted. Well, on my way to show the
puzzle to Williams, I stopped at the Brooklyn Eagle office to ask
Anthony Fiala, their artist and an old friend of mine, to touch it
up a bit for me. I could draw pretty well, but of course he knew
more about it than I did. He was so taken with the puzzle that
72 Penguin Book of Curious and Interesting Puzzles
he insisted on showing it first to the editor, then to the publisher,
and finally to the proprietor of the paper. They all wanted to
buy it, but I told them it was disposed of. Finally they proposed
that I should run a puzzle department for the Eagle; and before I
left them they had given me an order for $250 worth of copies
of the puzzle, and agreed to a salary of $50 a week for the
puzzle column.
The element of trickery always appealed to Loyd. He once gave a
display of mind-reading, using his son as stooge. His son appeared to
give correct answer after correct answer as Loyd held up a sequence
of cards behind his back, yet his son was only miming - Loyd
provided the answers himself by ventriloquism. It was said that a
family servant-girl had once left their service because she heard
'voices' every day in the parlour chimney.
Loyd was an excellent mimic, and enjoyed magic tricks and sleight
of hand and telling wonderful stories with which he amused his own
children - shades of Lewis Carroll.
He was also a self-taught wood-engraver and cartoonist, as well as
writer, publisher and editor. He once edited a mechanics trade paper,
and produced for a number of years Sam Loyd's Puzzle Magazine.
In between the trick donkeys and the vanishing Chinamen, and in
complete contrast, he produced in 1878 the '14-15' puzzle (problem 258),
which was the Rubik's Cube of the 1870s, and more. The craze swept
America, where employers posted notices forbidding employees to
play with the puzzle during office hours, and then crossed to Europe.
In Germany, Deputies in the Reichstag were observed huddled over
the little squares, in France it was called 'Ie Jeu de Taquin' (see p.
The Puzzles 73
64) and was described as a greater scourge than alcohol or tobacco.
Its spread was aided by the offer of a $1000 prize, a very large sum
in those days, for the first person to achieve a particular apparently
innocuous position. Loyd had first asked a New York newspaper
owner to put up the prize but he refused, and so Loyd offered it
himself, risking nothing because the puzzle was impossible - which
also meant that Loyd was not able to patent it.
According to Alain C. White, his friend and author of Sam Loyd
and His Chess Problems, 'Ideas came to him with great fecundity,
often too rapidly for him to analyse them completely. Yet his powers
for rapid analysis were almost unrivalled. He could see an idea from
many sides at once; first always from the point of view of a puzzle,
then from the humorous standpoint, finally from the artistic aspect.'
It is a curiosity, and a problem for psychologists to explain, that
someone so creative and so brilliant at chess-problem composition
should not have been a better over-the-board player or as good a
mathematician as Dudeney. Fortunately, in his puzzles his fecund
imagination, his ingenuity and his sense of humour were given full
reign.
After his death, his son, also named Sam, continued to produce
puzzles for newspapers, and also made collections of his father's
work. He lacked his father's talents, but possessed ample nerve,
exploiting the fact that they shared the same name to write as late as
1928, Sam Loyd and his Puzzles: An Autobiographical Review. He
had previously compiled, in 1914, Sam Loyd's Cyclopaedia of 5000
Puzzles, Tricks and Conundrums. The following puzzles are taken
from these two books.
This puzzle perfectly illustrates Loyd's ingenuity and humour, as well
as his ability to turn his puzzles into money. The theme is reminiscent
of the frog climbing out of the well.
253. 'Many years ago, when Barnum's Circus was of a truth "the
greatest show on earth", the famous showman got me to prepare for
him a series of prize puzzles for advertising purposes. They became
widely known as the Questions of the Sphinx, on account of the large
prizes offered to anyone who could master them.
'Barnum was particularly pleased with the problem of the cat and
dog race, letting it be known far and wide that on a certain day of
April he would give the answer and award the prizes, or, as he aptly
put it, "let the cat out of the bag, for the benefit of those most con-
cerned".
74 Penguin Book of Curious and Interesting Puzzles
'The wording of the puzzle was as follows:
'''A trained cat and dog run a race, one hundred feet straight-away
and return. The dog leaps three feet at each bound and the cat but
two, but then she makes three leaps to his two. Now, under those
circumstances, what are the possible outcomes of the race?"
'The fact that the answer was to be made public on the first of
April, and the sly reference to "letting the cat out of the bag", were
enough to intimate that the great showman had some funny answer
up his sleeve.'
Sam Loyd invented the cryptarithm, and it is appropriate that this
example should be a long division sum:
254. Can you restore the missing digits?
.9 ••
•• 4 •
• • 4 •
• • • •
'The archaeologist is examming a completed problem in long division,
engraved on a sandstone boulder. Due to weathering of the rock,
most of the figures are no longer legible. Fortunately, the eight legible
digits provide enough information to enable you to supply the missing
figures.
'It really looks as if there should be scores of correct answers, yet
The Puzzles 75
so far as 1 am aware, only one satisfactory restoration of the problem
has been suggested.'
255. How many acres are in the intenor triangular lake?
'I went to Lakewood the other day to attend an auction sale of some
land, but did not make any purchases on account of a peculiar
problem which developed. The land was advertised as shown in the
posters on the fence as 560 acres, including a triangular lake. The
three plots show the 560 acres without the lake, but since the lake
was included in the sale, I, as well as other would-be purchasers,
wished to know whether the lake area was really deducted from the
land.
'The auctioneer guaranteed 560 acres "more or less". This was not
satisfactory to the purchasers, so we left him arguing with katydids,
and shouting to the bullfrogs in the lake, which in reality was a
swamp.
'The question 1 ask our puzzlists is to determine how many acres
there be in that triangular lake, surrounded as shown by square plots
of 370, 116 and 74 acres. The problem is of peculiar interest to those
of a mathematical turn, in that it gives a positive and definite answer
76 Penguin Book of Curious and Interesting Puzzles
to a proposition which, according to usual methods, produces one of
those ever-decreasing, but never-ending decimal fractions.'
256. Rearrange the six pieces to make the best possible picture of a
horse.
'Many years ago, when I was returning from Europe in company
with Andrew G. Curtin, the famous war Governor of Pennsylvania
(returning from his post in Russia to seek nomination for president of
the United States) we discussed the curious White Horse monument
on Uffington Hill, Berkshire, England.
' If you know nothing about that weird relic of the early Saxons,
the accompanying sketch will afford an excellent idea of its appear-
ance. It represents the figure of a colossal white horse, several
hundred feet long, engraved on the side of the mountain about a
thousand feet above the level of the sea and easily seen from a
distance of some fifteen miles ... After the white horse had been
thoroughly discussed, the Governor banteringly exclaimed, "Now,
Loyd, there would be a capital subject for a puzzle."
The Puzzles 77
'Many a good puzzle idea has come from just such a tip. So, with
my scissors and a piece of silhouette paper, I speedily improvised the
accompanying figure of a horse.
'It would be a simple matter to improve the parts and general form
of the old horse, and I did modify it in the version which I afterwards
published, but somehow I love the old nag best as first devised, with
all its faults, so I now present it as it actually occurred ro me.
'The world has been moving rapidly during the last decade, and
puzzlers are much sharper than they used to be. In those days very
few, probably not one out of a thousand, actually mastered the
puzzle, so it will be a capital test of the acumen of the past compared
with that of the present generation ro see how many clever wits of
today can solve it.
'Trace an exact copy of the figure as shown. Cut out the six pieces
very carefully, then try to arrange them to make the best possible
figure of a horse. That is all there is to it, but the entire world
laughed for a year over the many grotesque representations of a horse
that can be made with those six pieces.
'I sold over one thousand million [sic] copies of "The Pony Puzzle".
This prompts me to say that whereas I have brought out many
puzzles, patented numerous inventions, and devoted much time and
money, to my sorrow, upon the "big things", more money is made
from little things like "The Pony Puzzle", which do not require a
five-dollar bill to promote and place on the market.'
257. How would you cut this gingerbread dog's head into two pieces
of the same shape?
78 Penguin Book of Curious and Interesting Puzzles
'Here is a practical problem in simple division calculated to baffle
some of our puzzlists. You see, Toodles has received the present of a
gingerbread dog's head and is told that she must divide it evenly with
her little brother. In her anxiety to be fair and equitable in the matter,
she wishes to discover some way to divide the cake into two pieces of
equal shape and size.
'How many of our clever puzzlists can come to her assistance by
showing how the dog's head may be divided?'
258. Slide the numbered blocks into serial order.
'Older inhabitants of Puzzleland will remember how in the seventies I
drove the entire world crazy with a little box of movable blocks
which became known as the "14-15 Puzzle". The fifteen blocks were
arranged in the square box in regular order, but with the 14 and 15
reversed as shown in the above illustration. The puzzle consisted of
moving the blocks about, one at a time, to bring them back to the
present position in every respect except that the error in the 14 and 15
was corrected.
'A prize of $1000, offered for the first correct solution to the
problem, has never been claimed, although there are thousands of
persons who say they performed the required feat.
'People became infatuated with the puzzle and ludicrous tales are
told of shopkeepers who neglected to open their stores; of a distin-
The Puzzles 79
guished clergyman who stood under a street lamp all through a
wintry night trying to recall the way he had performed the feat.
The mysterious feature of the puzzle is that none seem to be able
to remember tlie sequence of moves whereby they feel sure they
succeeded in solving the puzzle. Pilots are said to have wrecked
their ships, and engineers rush their trains past stations. A famous
Baltimore editor tells how he went for his noon lunch and was
discovered by his frantic staff long past midnight pushing little
pieces of pie around on a plate! Farmers are known to have deserted
their plows, and I have taken one such instance as an illustration
for the sketch.'
Loyd then gives three further puzzles, of which this is the first:
'Start again with the blocks as shown in the large illustration and
move them so as to get the numbers in regular order, but with the
vacant square at upper-left-hand corner instead of right-hand corner.
1 2 3
4 5 6 7
8 9 10 11
12 13 14 15
259. '''What is the age of that boy?" asked the conductor. Flattered
by this interest shown in his family affairs, the suburban resident re-
plied:
'''My son is five times as old as my daughter, and my wife is five
times as old as the son, and I am twice as old as my wife, whereas
grandmother, who is as old as all of us put together, is celebrating her
eighty-first birthday today."
'How old was the boy?'
260. An Odd Catch 'Ask your friends if they can write down five
odd figures that will add up to fourteen. It is really astonishing how
engrossed most people will get, and how much time they will spend
over this seemingly simple problem. You must be careful, however, to
say "figures" and not "numbers".'
261. Casey's Cow '''Some cows have more sense than the average
man," said Farmer Casey. "MyoId brindle was standing on a bridge
80 Penguin Book of Curious and Interesting Puzzles
the other day, five feet from the middle of the bridge, placidly looking
into the water. Suddenly she spied the lightning express, just twice the
length of the bridge away from the nearest end of the bridge, coming
toward her at a 90-mile an hour clip.
'''Without wasting a moment in idle speculation, the cow made a
dash toward the advancing train and saved herself by the narrow
margin of one foot. If she had followed the human instinct of running
away from the train at the same speed, three inches of her rear would
have been caught on the bridge!"
'What is the length of the bridge and the gait of Casey's cow?'
262. The Missing Link 'A farmer had six pieces of chain of five
links each, which he wanted made into an endless piece of thirty
links. If it costs eight cents to cut a link open and eighteen cents to
weld it again, and if a new endless chain could be bought for a dollar
and a half, how much would be saved by the cheapest method?'
263. Rearrange the eight pieces to form a perfect chessboard.
'In the history of France is told an amusing story of how the Dauphin
saved himself from an impending checkmate, while playing chess
with the Duke of Burgundy, by smashing the chessboard into eight
The Puzzles 81
pieces over the Duke's head. It is a story often quoted by chess
writers to prove that it is not always politic to play to win, and has
given rise to a strong line of attack in the game known as the King's
gambit.
'The smashing of the chessboard into eight pieces was the feature
which always struck my youthful fancy because it might possibly
contain the elements of an important problem. The restriction to
eight pieces does not give scope for great difficulty or variety, but not
feeling at liberty to depart from historical accuracy, I shall give our
puzzlists a simple little problem suitable for summer weather. Show
how to put the eight pieces together to form a perfect 8 x 8 checker-
board.
'The puzzle is a simple one, given to teach a valuable rule which
should be followed in the construction of puzzles of this kind. By
giving no two pieces the same shape, other ways of doing the puzzle
are prevented, and the feat is much more difficult of accomplish-
ment.'
264. 'An ancient problem, to be found in many old puzzle books,
concerns an army fifty miles long. As the army marches forward at a
constant rate, a courier starts at the rear of the army, rides forward to
deliver a message to the front, then returns to his position at the rear.
He arrives back exactly at the time that the army completed an
advance of fifty miles. How far altogether did the courier travel?
'If the army were stationary, he could clearly have to travel fifty
miles forward and the same distance back. But because the army is
advancing, he must go more than fifty miles to the front, and on his
return trip he will travel less than fifty miles because the rear of the
army is advancing towards him. It is assumed, of course, that the
courier always rides at a constant speed.
'A more difficult puzzle is created by the following extension of the
theme. A square army, fifty miles long by fifty miles wide, advances fifty
miles at a constant rate while a courier starts at the middle of the rear
and makes a complete circuit around the army and back to his starting
point. The courier's speed is constant, and he completes his circuit just
as the army completes its advance. How far does the courier travel?'
265. Tandem Bicycle 'Three men wish to go forty miles on a
tandem bicycle that will carry no more than two at a time while the
third man is walking. One man, call him A, walks at a rate of one
mile in ten minutes, B can walk a mile in fifteen minutes, and C can
walk a mile in twenty minutes. The bicycle travels at forty miles an
82 Penguin Book of Curious and Interesting Puzzles
hour regardless of which pair is riding it. What is the shortest time
for all three men to make the trip, assuming, of course, that they use
the most efficient method of combining walking and cycling?'
266. A Swiss Puzzle 'This pretty Swiss miss is extremely clever at
working geometrical cutting puzzles. She has discovered a way of
cutting the piece of red wall paper in her right hand into two pieces
that will fit together to form the Swiss flag she is holding in her left
hand. The white cross .in the centre-of the flag is actually a hole in the
paper. The cutting must follow the lines ruled on the paper.
'For a second puzzle, the Swiss girl asks you to cut the Rag in her
left hand into two pieces that will fit together to make a rectangle of
five-by-six units.
'Someone once asked the Swiss girl how to make a Maltese cross
and she replied, "Pull its tai!!'"
The Puzzles 83
267. DIssecting a Cross ~ I t is a remarkable fact that a mysterious
affinity, or relationship, can be shown to exist between all the ancient
signs and symbols, in that each one can be converted into another by
rearrangement of its dissected parts. Thus, a Swastika can be changed
into a square, the square into a cross, the cross into a triangle, etc.,
etc.
'One of these interesting transformations consists in dissecting the
Greek Cross, shown above, into four similar parts which may be
regrouped to present a square, with a central open space in the form
of a small Greek Cross. Work out the problem mentally before
applying your scissors.'
268. The Trapezoid Puzzle How can these five pieces be variously
assembled to form a square, a Greek Cross, a rhombus, a rectangle,
and a triangle? All five pieces must be used for each assembly.
84 Penguin Book of Curious and Interesting Puzzles
269. A Paradoxical Puzzle This puzzle perfectly illustrates Loyd's
ability to spot, in a familiar situation, what others had missed. This
diagram shows how an 8 x 8 square can be dissected and reassembled
to, apparently, gain an extra square, since the rectangle is 5 x 13.

• .,..!IIIi···.····
• I···
••••• 1I111i0.;: •••
DIE.... . ...
Loyd asks for another way to reassemble the same four pieces which
will lose a square.
270. Honest John 'Here is an extension of the "measuring" idea,
which you will find more elusive. I first published it in t900, as
follows:
'Honest John, the milkman, says that what he doesn't know about
milk is scarcely worth mentioning, but he was nearly flabbergasted the
other day when he got out on his route with his two to-galion cans full
of the lacteal fluid, but minus his measuring cans. Then along came
two customers, one with a five-quart pail and the other with a four-
quart pail, and they each demanded two quarts of milk in a hurry. In
filling the orders John proved himself considerable of a puzzler.
'To measure exactly two quarts of milk in each of those pails is a
measuring problem pure and simple, devoid of trick or device, but it
requires considerable cleverness to achieve the desired result with the
fewest number of pourings.'
271. Delicatessen Arithmetic 'Mrs Simpkins counted out the correct
amount of money and said to delicatessen Louis: "Give me a pound
and a half of bologna for boarders."
'Louis cut off a piece, weighed it, and remarked: "It weighs toc
over."
'''Then give me half of it, and the remainder of the money will buy
Sc worth of pickles," said Mrs Simpkins.
'How much did she expend on the bologna?'
The Puzzles 85
272. Carving a Doughnut 'The design shows the sort of doughnut
that buddies claim the Salvation Army lassies turned out "over there".
Whether ,intended or not, this particular doughnut makes an interest-
ing puzzle. Just draw a straight line across the doughnut to show how
many pieces you could produce with one straight cut.'
273. Popping the Question 'Danny went over to urge Kate to name
the day.
'''This is entirely unexpected," gasped the maiden; "but I will
marry you when the week after next is the week before last."
'''Had I received this promise yesterday," said Danny, "the waiting
would have been six days shorter."
'Can you tell on what day of the week Danny popped the ques-
tion?'
274. Make it Square 'This design [overleaf] contains exactly sixty-
four little squares, and the puzzle consists in showing how it may be
cut into the least possible number of pieces to make a large eight by
eight square, with the pattern preserved. How do you do it?'
86 Penguin Book of Curious and Interesting Puzzles
275. Everything Free 'A little girl visited the food show and ate
seventeen different kinds of breakfast food and gathered 10 pounds of
sample packages. Then she stepped on the free weighing machine and
found that her weight had increased 10 per cent, whereas if she had
eaten twice as much breakfast food the gain would have been 11 per
cent. What was her weight when she arrived at the food show?'
A Revolutionary Rebus Loyd published hundreds of rebuses and
other simple puzzles. This example is included by way of illustration,
because of its ingenuity. As Loyd presented it:
276. 'One of the incidents leading to the Revolutionary War, especi-
ally interesting to young students, is represented by that monogram.
What was the historical event?'
The Longest Queen's Tour This puzzle appeared in Le Sphinx in
March 1867, and was described by Loyd as the best of Its kind he had
ever produced and as 'the most difficult puzzle extant':
The Puzzles 87
277. The problem is simply to 'Place the Queen on [a chessboard)
and pass her over the entire sixty-four squares and back again to
point of beginning in fourteen moves.'
278. The Same Again, Almost 'Pass the Queen over the centre
points of all the squares in fourteen straight moves, returning to the
starting point.'
('Straight moves' are not limited to the ordinary Queen moves in
chess. Any move in a straight line will do.)
279. Dissecting the Chessboard 'As interesting as any of the dissec-
tion puzzles based on the chessboard, is the one which asks:
'What is the maximum number of pieces into which a chessboard
can be divided without any two of the pieces being exactly alike?
'Of course, the first step is to mark off a single white square and a single
black one. These, alike in size, are dissimilar in colour. There we have the
idea. Two pieces may be alike in form and size yet dissimilar in number or
arrangement of white and black squares. Pieces are considered unalike if
dissimilar in any respect. It is a pretty problem and not too difficult.'
280. Knight Dissection 'Divide the chessboard [overleaf), on the
lines, into four exactly equal parts, so that there shall be one of the
Knights in each of the parts.'
88 Penguin Book of Curious and Interesting Puzzles
281. From the Start. In the first number of the American Chess
Journal Loyd introduced the series of chess puzzles based on the
ordinary line-up of the pieces which has since become so famous.
'It will not be amiss,' he wrote, 'to have a little impromptu
exhibition, bearing upon conditional positions produced from the
position of the forces as arranged for actual play. I find two by
Breitenfeld, one by Max Lange, some from "Sissa", Dr Moore, etc.,
but as all can be solved in less moves than intended by the authors, I
give them under one heading, without authorship, and I have thrown
in a few similar ideas that occurred to me, elucidated in a sketch.'
The Puzzles 89
(a) If both parties move the same moves, how can the first player
mate in four moves?
(b) If both parties make the same moves, how can the first player self-
mate on the eighth move?
(c) Fmd how discovered checkmate can be effected in four moves.
(d) Find how a stalemate might result in ten moves.
(e) Find a game wherein perpetual check can be forced from the third
move.
Henry Dudeney
Henry Dudeney was born in the village of Mayfield in Sussex. His
paternal grandfather was a self-taught mathematician and astronomer
who started as a shepherd and raised himself to the position of
schoolmaster in the town of Lewes. Dudeney's father was also a
schoolmaster, but Dudeney himself did not go to college and was also
a self-taught mathematician.
He enjoyed games and was a good chess player, though, like Sam
Loyd, he was a better problemist, as might be expected. He also
played croquet, a game that might have been designed for puzzlists,
and entertained children with displays of magic and legerdemain.
He started composing puzzles under the pseudonym 'Sphinx', and
for a while he collaborated with Sam Loyd. When their collaboration
ended, Dudeney published under his own name in a variety of
magazines: the Strand, Cassell's, the Queen, TIt-Bits, the Weekly
Dispatch and Blighty. So, like modern television stand-up comedy
writers, he had to keep up a constant flow of ideas. Sam Loyd was in
the same position. This makes it all the more astonishing that their
levels were so consistently high. Loyd showed greater ingenuity in
exploiting his puzzles, especially for advertising. He had an uncanny
knack for appealing to the public. Yet Dudeney was much the better
mathematician, and his puzzles are more mathematically sophisti-
cated, without requiring any mathematics beyond the most
elementary. *
Dudeney was interested in the psychology of puzzles and puzzle-
solving. In the original preface to A Puzzle-l\1i,ze he asserted that 'The
fact is that our lives are largely spent in solving puzzles; for what is a
puzzle but a perplexing question? And from our childhood upwards
we are perpetually asking questions or trying to answer them.'
* I am mdebted to Martin Gardner's Introduction to 536 Puzzles and Curious
Problems for some of thiS mformatlon.
90 Penguin Book of Curious and Interesting Puzzles
But he was also a man of his age. In the same preface he remarks
that 'The solving of puzzles consists merely in the employment of our
reasoning faculties, and our mental hospitals are built expressly for
those unfortunate people who cannot solve puzzles.' Elsewhere he
remarks that 'The history of [mathematical puzzles) entails nothing
short of the actual story of the beginnings and development of exact
thinking in man.'
Half a century later we are more aware of the roles of insight and
imagination, and the 'Aha!' response, which are more than the
exercise of logic or reason. Dudeney also supposed that puzzles had
great value in training the mind, a natural assumption in the days
when many educational theorists still believed in the idea of mental
training. We would put that rather differently: it is mathematics
teachers today who most exploit puzzles and mathematical recreations
to entice their pupils and to illuminate mathematical ideas.
Fortunately, just as an artist may have a naive theory of their own
art, so Dudeney's puzzles are not limited by his own interpretations
of them. They exhibit a wealth of imagination and ingenuity, even
artistry ...
The following puzzles are selected from The Canterbury Puzzles
(1907), Amusements in Mathematics (1917) and Modern Puzzles
(1926). The title puzzles of The Canterbury Puzzles are a sequence of
problems proposed by 'A chance-gathered company of pilgrims, on
their way to the shrine of Saint Thomas a Becket at Canterbury, met
at the Tabard Inn, later called the Talbot, in Southwark [whose) host
proposed that they should beguile the ride by each telling a tale to his
fellow-pilgrims. '
282. The Haberdasher's Puzzle 'Many attempts were made to
induce the Haberdasher, who was of the party, to propound a puzzle
of some kind, but for a long time without success. At last, at one of
the Pilgrim's stopping-places, he said that he would show them
something that would "put their brains into a twist like unto a bell-
rope". As a matter of fact, he was really playing off a practical joke
on the company, for he was quite ignorant of any answer to the
puzzle that he set them. He produced a piece of cloth in the shape of
a perfect equilateral triangle, as shown in the illustration, and said,
"Be there any among ye full wise in the true cutting of cloth? I trow
not. Every man to his trade, and the scholar may learn from the
varlet and the wise man from the fool. Show me, then, if ye can, in
what manner this piece of cloth may be cut into four several pieces
The Puzzles 91
that may be put together to make a perfect square."
'Now some of the more learned of the company found a way of
doing it in five pieces, but not in four. But when they pressed the
Haberdasher for the correct answer he was forced to admit, after
much beating about the bush, that he knew no way of doing it in any
number of pieces. "By Saint Francis," saith he, "any knave can make
a riddle methinks, but it is for them that may to rede it right." For
this he narrowly escaped a sound beating. But the curious point of
the puzzle is that I have found that the feat really may be performed
in so few as four pieces, and without turning over any piece when
placing them together. The method of doing this is subtle, but I think
the reader wlil find the problem a most interesting one.'
283. The Spider and the Fly 'Inside a rectangular room, measuring
30 feet in length and 12 feet in width and height, a spider is at a point
in the middle of one of the end walls, 1 foot from the ceiling, as at A;
and a fly is on the opposite wall, 1 foot from the floor in the centre,
92 Penguin Book of Curious and Interesting Puzzles
...
as shown at B. What is the shortest distance that the spider must
crawl in order to reach the fly, which remains stationary? Of course
the spider never drops or uses its web, but crawls fairly.'
.,
-
140 loc
)(1
I
The Puzzles 93
284. Catching the Hogs 'In the Illustration Hendrick and Katrun
are seen engaged in the exhilarating sport of attempting the capture
of a couple of hogs.
'Why did they fail?
'Strange as it may seem, a complete answer is afforded in the little
puzzle game that I will now explain.'
[Dudeney instructs the reader to represent the Dutchman and his
wife, and the two hogs, by four counters, on squared paper.]
'The first player moves the Dutchman and his wife one square each
in any direction (but not diagonally), and then the second player
moves both pigs one square each (not diagonally); and so on, in
turns, until Hendrick catches one hog and Katrun the other.
'This you will find would be absurdly easy if the hogs moved first,
but this IS just what Dutch pigs will never do.'
285. Making a Flag 'A good dissection puzzle in so few as two
pieces is rather a rarity, so perhaps readers will be interested in the
following. The diagram represents a piece of bunting, and it is
required to cut it into two pieces (without any waste) that will fit
together and form a perfectly square flag, with the four roses sym-
metrically placed. This would be easy enough if it were not for the
four roses, as we should merely have to cut from A to B and insert
the piece at the bottom of the flag. But we are not allowed to cut
through any of the roses, and. therein lies the difficulty of the puzzle.'
94 Penguin Book of Curious and Interesting Puzzles
286. Bridging the Ditch As Dudeney describes it, this plan is of a
ditch which is 10 feet wide, and filled with water. The King's Jester
desperately requires to cross the ditch, but he cannot swim. The only
equipment he can find is a heap of eight planks, each of which is no
more than 9 feet long.
How can he cross the ditch to safety?
Lady Isabel's Casket This puzzle is the first appearance of the idea
of a 'squared square', that is, a square dissected into distinct smaller
squares, though Dudeney has to resort to a narrow rectangular strip
to fill a portion of the surface.
287. 'Sir Hugh's young kinswoman and ward, Lady Isabel de Fitz-
arnulph, was known far and wide as "Isabel the Fair". Amongst her
treasures was a casket, the top of which was perfectly square in
shape. It was inlaid with pieces of wood, and a strip of gold ten
inches long by a quarter of an inch wide.
'When young men sued for the hand of Lady Isabel, Sir Hugh
promised his consent to the one who would tell him the dimensions
of the top of the box from these facts alone: that there was a
rectangular strip of gold, ten inches by 1I4-inch; and the rest of the
surface was exactly inlaid with pieces of wood, each piece being a
perfect square, and no two pieces the same size. Many young men
failed, but one at length succeeded. The puzzle is not an easy one, but
the dimensions of that strip of gold, combined with those other
conditions, absolutely determine the size of the top of the casket.'
The Puzzles 95
288. The Fly and the Cars A road is 300 miles long. A car, A, starts
at noon from one end and goes throughout at 50 miles an hour, and
at the same time another car, B, going uniformly at 100 miles an
hour, starts from the other end together with a fly travelling at 150
miles an hour. When the fly meets car A, it immediately turns and
flies towards B.
(1) When does the fly meet B?
The fly then turns towards A and continues flying backwards and
forwards between A and B.
(2) When wiII the fly be crushed between the cars if they collide and it
does not get out of the way?
289. Crossing the Moat 'I [the King's Jester, still adventuring) was
now face to face with the castle moat, which was, indeed, very wide
and very deep. Alas! I could not swim, and my chance of escape
seemed of a truth hopeless, as, doubtless, it would have been had 1
not espied a boat tied to a wall by a rope. But after 1 had got into it 1
did find that the oars had been taken away, and that there was
nothing that 1 could use to row me across. When 1 had untied the
rope and pushed off upon the water the boat lay quite still, there
being no stream or current to help me. How, then, did 1 yet take the
boat across the moat?'
290. The Crescent and the Cross 'When Sir Hugh's kinsman, Sir
John de Collingham, came back from the Holy Land, he brought
96 Penguin Book of Curious and Interesting Puzzles
with him a flag bearing the sign of a crescent, as shown in the
illustration. It was noticed thai: de Fortibus spent much time in
examining this crescent and comparing it with the cross borne by the
Crusaders on their own banners ... [He] explained that the crescent
in one banner might be cut into pieces that would exactly form the
perfect cross in the other. It is certainly rather curious; and I shall
show how the conversion from crescent to cross may be made in ten
pieces, using every part of the crescent. The flag was alike on both
sides, so pieces may be turned over where required.'
291. The Riddle of St Edmondsbury '''It used to be told at St
Edmondsbury," said Father Peter on one occasion, "that many years
ago they were so overrun with mice that the good abbot gave orders
that all the cats from the country round should be obtained to
exterminate the vermin. A record was kept, and at the end of the year
it was found that every cat had killed an equal number of mice, and
the total was exactly 1,111,111 mice. How many cats do you suppose
there were?'
292. The Cigar Puzzle 'Two men are seated at a square-topped table.
One places an ordinary cigar (flat at one end, pointed at the other) on
the table, then the other does the same, and so on alternately, a
condition being that no cigar shall touch another. Which player should
succeed in placing the last cigar, assuming that they each will play in
the best possible manner? The size of the table top and the size of the
cigar are not given, but in order to exclude the ridiculous answer that
the table might be so diminutive as only to take one cigar, we will say
that the table must not be less than 2 feet square and the cigar not more
than 4t inches long. With those restrictions you may take any dimen-
sions you like. Of course we assume that all the cigars are exactly alike
in every respect. Should the first player, or the second player, win?'
293. The Damaged Measure 'Here is a new puzzle that is interest-
ing, and it reminds me, though it is really very different, of the
classical problem by Bachet concerning the weight that was broken in
pieces which would then allow of any weight in pounds being
determined from one pound up to a total weight of all the pieces. In
the present case a man has a yard-stick from which 3 inches have
been broken off, so that it is only 33 inches in length.
'Some of the graduation marks are also obliterated, so that only
eight of these marks are legible; yet he is able to measure any given
The Puzzles 97
number of inches from 1 inch up to 33 inches. Where are these marks
placed?'
294. Exploring the Desert 'Nine travellers, each possessing a motor-
car, meet on the eastern edge of a desert. They wish to explore the
Intenor, always going due west. Each car can travel forty miles on the
contents of the engine tank, which holds a gallon of petrol, and each
can carry nine extra gallon tins of petrol and no more. Unopened tins
can alone be transferred from car to car. What is the greatest distance
to which they can enter the desert without making any depots for
petrol for the return journey?'
295. A Puzzle with Pawns 'Place two pawns in the middle of the
chessboard, one at Q4 and the other at KS. Now, place the remaining
fourteen pawns (sixteen in all), so that no three shall be in a straight
line in any possible direction.
'Note that I purposely do not say queens, because by the words
"any possible direction" I go beyond attacks on diagonals. The
pawns must be regarded as mere points in space - at the centres of
the squares.'
296. The Game of Bandy-ball Bandy-ball, cambuc, or goff (the
game so well known today by the name of golf), is of great antiquity,
and was a special favourite at Solvemhall Castle. Sir Hugh de Forti-
98 Penguin Book of Curious and Interesting Puzzles
bus was himself a master of the game, and he once proposed this
question.
'They had nine holes, 300, 250, 200, 325, 275, 350, 225, 375, and
400 yards apart. If a man could always strike the ball in a perfectly
straight line and send it exactly one of two distances, so that it would
either go towards the hole, pass over it, or drop into it, what would
the two distances be that would carry him in the least number of
strokes round the whole course?
'Two very good distances are 125 and 75, which carry you round in
twenty-eight strokes, but this is not the correct answer. Can the
reader get round in fewer strokes with two other distances?'
297. The Noble Demoiselle 'Seated one night in the hall of the
castle, Sir Hugh desired the company to fill their cups and listen
while he told the tale of his adventure as a youth in rescuing from
captivity a noble demoiselle who was languishing in the dungeon of a
castle belonging to his father's greatest enemy ... Sir Hugh produced
a plan of the thirty-five cells in the dungeon and asked his companions
to discover the particular cell that the demoiselle occupied. He said
that if you started at one of the outside cells and passed through
every doorway once, and once only, you were bound to end at the
cell that was sought. Can you find the cell? Unless you start at the
correct outside cell it is impossible to pass through all the doorways
once and once only.'
The Puzzles 99
298. The Trusses of Hay 'Farmer Tomkins had five trusses of hay,
which he told his man Hodge to weigh before delivering them to a
customer. The stupid fellow weighed them two at a time in all
possible ways, and .informed his master that the weights in pounds
were 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121. Now, how
was Farmer Tompkins to find out from these figures how much every
one of the five trusses weighed singly? The reader may at first think
that he ought to be told "which pair is which pair" or something of
that sort, but it IS quite unnecessary. Can you give the five correct
weights?'
299. Another Joiner's Problem 'A joiner had two pieces of wood of
the shapes and relative proportions shown in the diagram. He wished
to cut them into as few pieces as possible so that they could be fitted
together, without waste, to form a perfectly square table-top. How
should he have done it? There is no necessity to give measurements,
for if the smaller piece (which is half a square) be made a little too
large or small, It will not affect the method of solution.'
100 Penguin Book of Curious and Interesting Puzzles
300. The Rook's Tour 'The puzzle is to move the single rook over
the whole board, so that it shall visit every square of the board once,
and only once, and end its tour on the square on which it starts. You
have to do this in as few moves as possible.'
301. The Five Pennies 'Every reader knows how to place four
pennies so that they all touch one another. Place three in the form of
a triangle, and lay the fourth penny on top in the centre. Now try to
do the same with five pennies - place them so that every penny shall
touch every other penny.'
302. De Morgan and Another 'Augustus de Morgan, the mathemati-
cian, who died in 1871, used to boast that he was x years old in the
year x
2
• My living friend, jasper jenkins, wishing to improve on this,
tells me that he was a
2
+ b
l
in a
4
+ b
4
; that he was 2m in the year
2m
2
; and that he was 3n years old in the year 3n
4
• Can you give the
years in which De Morgan and jenkins respectively, were born?'
303. Sum Equals Product '''This is a curious thing," a man said to
me. "There are two numbers whose sum equals their product. They
are 2 and 2, for if you add them or multiply them, the result is 4."
Then he tripped badly, for he added, "These are, 1 find, the only two
numbers that have this property."
'I asked him to write down any number, as large as he liked, and 1
The Puzzles 101
would immediately give him another that would give a like result by
addition or multiplication.'
What was Dudeney's method of doing this?
Water, Gas and Electricity This puzzle illustrates the difficulty of
deciding which puzzles were invented by Dudeney and Loyd and
which were borrowed from each other, or from older sources. Loyd
claimed that he invented this puzzle about 1903, Dudeney thinks
otherwise. I am inclined to think that such a puzzle - of obvious
practical relevance - ought to be old, but if so then it is surprising
that it does not appear in the commonest Victorian puzzle books.
304. 'There are some half-dozen puzzles, as old as the hills, that are
perpetually cropping up, and there is hardly a month in the year that
does not bring inquiries as to their solution. Occasionally one of
these, that one had thought was an extinct volcano, bursts into
eruption in a surprising manner. I have received an extraordinary
number of letters respecting the ancient puzzle that I have called
"Water, Gas and Electricity". It is much older than electric lighting,
or even gas, but the new dress brings it up to date. The puzzle is to
lay on water, gas, and electricity, from W, G and E, to each of the
three houses, A, Band C, without any pipe crossing another.'
305. The Six Pennies Lay six pennies on the table, and then arrange
them as shown overleaf, so that a seventh would fit exactly into the
central space. You are not allowed the use of a ruler or any other
measuring device, just the six pennies.
102 Penguin Book of Curious and Interesting Puzzles
306. Placing Halfpennies 'Here is an interesting little puzzle sug-
gested to me by Mr W. T. Whyte. Mark off on a sheet of paper a
rectangular space 5 inches by 3 inches, and then find the greatest
number of halfpennies that can be placed within the enclosure under
the following conditions. A halfpenny is exactly an inch in diameter.
Place your first halfpenny where you like, then place your second
coin at exactly the distance of an inch from the first, the third an inch
distance from the second, and so on. No halfpenny may touch
another halfpenny or cross the boundary. Our illustration will make
the matter perfectly clear. No.2 coin is an inch from No.1; No.3 an
inch from No.2; No.4 an inch from No.3; but after No. 10 is placed
we can go no further in this attempt. Yet several more halfpennies
might have been got in. How many can the reader place?'
The Puzzles 103
307. The Bun Puzzle 'The three circles represent three buns, and it
is simply required to show how these may be equally divided among
four boys. The buns must be regarded as of equal thickness through-
out and of equal thickness to each other. Of course, they must be cut
into as few pieces as possible. To simplify it I will state the rather
surprising fact that only five pieces are necessary, from which it will
be seen that one boy gets his share in two pieces and the other three
receive theirs in a single piece. I am aware that this statement "gives
away" the puzzle, but it should not destroy its interest to those who
like to discover the "reason why".'
308. The Cardboard Chain 'Can you cut this chain out of a piece
of cardboard without any join whatsoever? Every link is solid, without
its having been split and afterwards joined at any place. It is an
interesting old puzzle that I learnt as a child, but I have no knowledge
as to its inventor.'

309. The Two Horseshoes 'Why horseshoes should be
"lucky" is one of those things which no man can understand. It is a
very old superstition, and John Aubrey (1626-1700) says, "Most
houses at the West End of London have a horseshoe on the threshold."
In Monmouth Street there were seventeen in 1813 and seven so late as
1855. Even Lord Nelson had one nailed to the mast of the ship
Victory. Today we find it more conducive to "good luck" to see that
they are securely nailed on the feet of the horse we are about to drive.
'Nevertheless, so far as the horseshoe, like the Swastika and other
104 Penguin Book of Curious and Interesting Puzzles
emblems that 1 have had occasion at times to deal with, has served to
symbolize health, prosperity, and goodwill towards men, we may
well treat it with a certain amount of respectful interest. May there
not, moreover, be some esoteric or lost mathematical mystery con-
cealed in the form of a horseshoe? 1 have been looking into this
matter, and 1 wish to draw my readers' attention to the very remark-
able fact that the pair of horseshoes shown in my illustration are
related in a striking and beautiful manner to the circle, which is the
symbol of eternity. 1 present this fact in the form of a simple problem,
so that it may be seen how subtly this relation has been concealed for
ages and ages. My readers will, 1 know, be pleased when they find the
key to the mystery.
'Cut out the two horseshoes carefully round the outline and then
cut them into four pieces, all different in shape, that will fit together
and form a perfect circle. Each shoe must be cut into two pieces and
all the part of the horse's hoof contained within the outline is to be
used and regarded as part of the area.'
310. The Table-Top and the Stools 'I have frequently had occasion
to show that the published answers to a great many of the oldest and
most widely known puzzles are either quite incorrect or capable of
improvement. 1 propose to consider the old poser of the table-top and
stools that most of my readers have probably seen in some form or
another in books compiled for the recreation of childhood.
'The story is told that an economical and ingenious schoolmaster
once wished to convert a circular table-top, for which he had no use,
into seats for two oval stools, each with a hand-hole in the centre. He
instructed the carpenter to make the cuts as in the illustration and
then join the eight pieces together in the manner shown. So impressed
was he with the ingenuity of his performance that he set the puzzle to
his geometry class as a little study in dissection. But the remainder of
The Puzzles 105
the story has never been published, because, so it is said, it was a
characteristic of the principals of academies that they would never
admit that they could err. I get my information from a descendant of
the original boy who had most reason to be interested in the matter.
'The clever youth suggested modestly to the master that the hand-
holes were too big, and that a small boy might perhaps fall through
them. He therefore proposed another way of making the cuts that
would get over this objection. For his impertinence he received such
severe chastisement that he became convinced that the larger the
hand-hole in the stools the more comfortable might they be.
'Now what was the method the boy proposed?
'Can you show how the circular table-top may be cut into eight
pieces that will fit together and form two oval seats for stools (each
of exactly the same size and shape) and each having similar hand-
holes of smaller dimensions than in the case shown above? Of course,
all the wood must be used.'
Send More Money Loyd was the first inventor of the cryptarithm,
in which some or all of the digits in a sum are deleted, and the sum
has to be reconstructed, but Dudeney first replaced the mi!>sing digits
with letters to make a meaningful message - he called it Verbal
Arithmetic - a rare example of Dudeney hitting upon a popular point
that Loyd missed.
SEND
MORE
MONEY
311. This is a correct addition sum, in which each different letter
stands for a different digit, zero possibly included. What is the
original sum?
106 Penguin Book of Curious and Interesting Puzzles
The Eight Spiders Just as Dudeney and Loyd used many old puzzles,
often adding new ideas of their own, so their puzzles have been
exploited by others. The problem of the spider and the fly (problem
283 above) has been especially fruitful in variations. This one is due
to Maurice Kraitchik.
312. 'An honourable family of spiders, consisting of a wise mother
and eight husky youngsters, were perched on the wall at one end of a
rectangular room. Food being scarce, owing to the Second World
War, they were grumbling, when an enormous fly landed unnoticed
on the opposite wall. If Euclid could have been summoned from his
grave (location, alas, unknown), he would have been able to show
that both the hunters and the prey were in the vertical plane bisecting
the two opposite walls, the spiders eighty inches above the centre and
the fly eighty inches below.
'Suddenly one young spider shouted with glee. "Mamma! Look!
There's a fly! Let's catch him and eat him!"
'''There are four ways to reach the fly. Which shall we take?"
came the eager query from another.
'''You have forgotten your Euclid, my darling. There are eight
ways to reach the fly. Each of you take a different path, without
using any other means of conveyance than your God-given legs.
Whoever reaches the goal first shall be rewarded with the largest
portion of the prey."
'At the signal given by the mother the eight spiders shot out in
eight different directions at a speed of 0.65 mile per hour. At the end
of W seconds they simultaneously converged on the fly, but found no
need of attacking it since its heart had given way at the sight of
enemies on all sides.
'What are the dimensions of the room?'
•
313. The Magic Hexagon How can the numbers 1 to 19 be placed
in the cells of this hexagon so that all fifteen sums, five each in each
of three directions, are equal?
In companson to normal magic squares, which sum in two directions
plus, more often than not, along some diagonals, it might be expected
that there would be fewer magic hexagons, summing in three direc-
tions. Yet it is surprising that there is just one unique solution to this
problem, apart from reflections and rotations, and no solutions at all
for any other size of hexagon.
The Puzzles 107
314. The Boat in the Bath Tommy was floating a boat in a tub of
water. The boat was initially loaded with a small metal cannon, but
then the cannon fell into the water and sank to the bottom, leaving
the boat floating as before. No water got into the boat while this
happened.
Did the level of the water rise, fall, or stay the same, as a result of
the cannon falling overboard?
Arithmorems This is the name given to a simple but elegant type of
puzzle, reminiscent of the 'Nuns Puzzle'.
315. How can the digits 1 to 9 be placed in these circles, so that each
side of the triangle sums to 20? How can the sum be made to equal
17?
108 Penguin Book of Curious and Interesting Puzzles
316. The Axle Poser Why does the front axle of a cart usually wear
out faster than the back?
317. Cutting the Cube It is easy to cut a cube into twenty-seven
smaller cubes by slicing it twice vertically, twice horizontally and
twice from back to front. This makes a total of six slices.
Suppose, however, that having made one or more slices you are
allowed to rearrange the pieces as you choose, before making the
next slice. Is it now possible to dissect the cube into twenty-seven
smaller cubes in only five slices?
318. Counting To and Fro Mike was standing in the doorway of
his house, ,counting the people passing in both directions. Tom was
walking up and down the road, counting all those he passed, in either
direction. After an hour, they met and compared the number they
had counted. Who had counted most?
319. A Terminating Division 'In the terminating division shown
below, the dots represent unknown digits, and in the entire division
the location of only seven digits 7 are known. However, it is also
possible that a dot represents a digit 7.'
•••• 7 •• 1 •••••• 7 ••••••••• 1 ••• 7 ••••••
• • • • • • •
• • • • • • • •
• • • • • • • •
• • • • • • • •
• • • • • • •
• • • • • • •
• • • • • • •
• ••••• 7 •
· . .7. . .
• • • • • • • •
• • • • • • • •
• • • • • • • •
• • • • • • •
• 7 •••••
. .7. . . .
• • • • • • •
• • • • • • •
•
Reconstruct the completed sum.
The Puzzles 109
320. The Overhanging Bricks (1) A large number of identical rec-
tangular bricks are piled in a vertical column, so that each brick is
immediately over the brick below. What is the greatest overhang that
can be achieved by sliding the bricks over each other, parallel to their
longest sides?
321. The Overhanging Bricks (2) With the same supply of bricks
and a table-top, what is the greatest projection possible, over the edge
of the table, using only four bricks?
The bricks may be arranged in any manner, but the maximum
projection is measured as the maximum distance from the edge of the
table to the end of a brick.
322. A Leap in Age The day before yesterday I was 13 years old.
Next year I shall be old enough to get married. When is my birthday,
and what is the date today?
323. The Cylindrical Hole A hole 6 inches long is drilled through the
centre of a solid sphere. What is the volume of the sphere remaining?
324. The Lost Pound Three diners on finishing their meal are
presented with a bill for £30, which they agree to split between them.
They each gave the waiter £10, not knowing that the waiter had
rechecked the bill and found that it was only £25. At this point it
occurred to the waiter that £5 would not divide equally between the
three, and anyway they did not know that there had been a mistake.
So he returned to the table, apologized and gave each diner £1,
keeping the other £2 for himself.
The diners have each now paid £9, making £27 in all, and the
waiter has £2 in his pocket, a total of £29. Yet they originally gave
the waiter £30. Where has the missing pound gone?
325. The Submerged Balance A lump of lead is being weighed on a
balance, placed on the Rat bottom of a basin, against several iron
weights. When the stone and weights balance exactly, the balance is
submerged by filling the basin with water.
Will it stay balanced? If not, which way will it tip?
326. 'This diagram [overleaf] shows a cube with a piece cut off. Your
problem is this: can you tell from this diagram if the slice ABCD
could be a Rat slice? That is, could the points, A, B, C and D lie in a
plane?'
110 Penguin Book of Curious and Interesting Puzzles
327. The neighbouring countries of Mona and Monia were both
jealous of each other's economic success, so when the government of
Moria announced that in future a Monia dollar would be worth only
90 Moria cents, the government of Moma retaliated by announcing
that in future each Moria dollar would, likewise, be worth only 90
Monia cents.
At once, a bright young spark who lived in Monia, near the
border, bought a 10 cent doughnut with his Monia dollar, exchanged
his 90 cents for a Moria dollar, crossed into Moria and bought
himself a soft drink for 10 cents. He then exchanged his 90 Moria
cents for a Monia dollar, leaving him with the money he started with.
Who paid for the doughnut and the soft drmks?
328. The Matching Birthdays How many people must be gathered
together in the same room, before you can be certain that there is a
greater than 50150 chance that at least two of them have the same birth-
day?
The surprising result was first noted by the mathematician Harold
Davenport.
329. Mastering the Masters Mary was a delightful child with a
precocious mterest in chess, so she ",as delighted when her father
arranged for two chess masters to visit her house. She was even more
pleased when they both agreed to playa game with her, and she was
ecstatic when she won one of the games, as conclusively as she lost
the other. Given that Mary was actually too weak to beat either of
the masters in a million years, what happened?
The Puzzles 111
Hubert Phillips
Hubert Phillips was a prolific composer of all kinds of puzzles who
had some strikingly original ideas, as well as producing many varia-
tions in particular themes, especially inference and deduction.
He often wrote under the pseudonyms of 'Caliban' in the New
Statesman, and 'Dogberry' in the News Chronicle. He was also editor
of British Bridge World and twice captain of England at contract
bridge, a humourist and the creator of the Inspector Playfair detection
mysteries.
In his mathematical puzzles he collaborated with his friend Sydney
Shovelton, and others. In the Introduction to The Sphinx Problem
Book he explained his conception of a good inferential-mathematical
puzzle: in particular, 'the line of approach must be as well concealed
as possible. I have put a good deal of thought into the construction of
problems which at first blush appear to be insoluble, through inad-
equacy of the data. The invention of such exercises, and the solving
of them, both give a great deal of pleasure, since their construction
can involve - and in my view should have reference to - principles of
artistry which embody an aesthetic of their own.'
In Question Time he also commented on his criteria for a good puzzle:
'Does its statement involve, not only the labour of working out the
answer (which for many has a very slight appeal) but also the excitement
of first discovering how the answer is to be arrived at? My main
pleasure, in constructing puzzles, lies in seeking to provide this "kick".'
I am reminded of the great mathematician G. H. Hardy's talking in
his Mathematician's Apology of 'the puzzle columns in the popular
newspapers. Nearly all their immense popularity is a tribute to the
drawing power of rudimentary mathematics, and the better makers of
puzzles, such as Dudeney or "Caliban", use very little else ... what the
public wants is a little intellectual "kick", and nothing else has quite
the kick of mathematics.'
The first two problems are based on an idea which first occurs in his
'Problems for Young Mathematicians' in The Playtime Omnibus:
330. What Colour was the Bear? A man out hunting, spotted a bear
due east. Taken by surprise, he ran directly north, and turned to see
that the bear had not moved. Steadying himself, he took aim and shot
it, by aiming due south. What colour was the bear?
112 Penguin Book of Curious and Interesting Puzzles
331. A Roundabout Journey Mrs Agabegyun left her house one
morning, and walked 5 miles south. She then turned east and walked
another 5 miles. Finally she turned again and walked 5 miles due
north, arriving back at her house.
Where does she live?
Phillips also originated the 'liars and truth-tellers' theme, which has
subsequently lent itself to endless variation:
332. Red and Blue The island of Ko is inhabited by three different
races - the Blacks, who invariably tell the truth; the Whites, who
invariably lie; and the Muddleds, who tell the truth and lie alternately
(though one cannot tell, in talking to a Muddled, whether his first
remark is truthful or the reverse).
'In a certain school in Ko, a Black, a White and a Muddled were
sitting side by side - in what order is not known. An inspector came in
carrying a number of cards, some of them red and some blue. Taking
a card at random, he asked each of the youngsters in turn: "What
colour is this?" Then, taking a second card, he put to each of them in
the same order, the same question. The six answers he received were:
(1) Blue, (2) Blue, (3) Red, (4) Red, (5) Blue, (6) Blue.'
What colour were the cards chosen by the inspector?
The next puzzle has also led to many variants, one of which follows im-
mediately:
333. 'Two schoolboys were playing on the toolshed roof. Something
gave way, and they were precipitated, through the roof, on to the
Roor below.
'When they picked themselves up, the face of one was covered with
grime. The other's face was quite clean. Yet it was the boy with the
clean face who at once went off and washed .
• How is this to be explained?'
334. The Three Wise Men Three Wise Men were taking a nap
when a practical joker marked a cross on the forehead of each, with
charcoal. The joker then hid behind a pillar and yelled loudly. At
once they awoke and each started laughing at the plight of the others,
until suddenly one of them stopped laughing and felt his own fore-
head, having realized that he was a victim of the same trick.
How did he draw this conclusion?
The Puzzles 113
335. The Ship's Ladder 'The good ship Potiphar lay at anchor in
Portsmouth Harbour. An interested spectator observed that a ladder
was dangling from her deck; that the bottom four rungs of the ladder
were submerged; that each rung was two inches wide and that the
rungs were eleven inches apart. The tide was rising at the rate of
eighteen inches per hour .
• At the end of two hours, how many rungs would be submerged?'
The next problem is a slight variant only of problem 521 in Dudeney's
536 Puzzles and Curious Problems. It originally appeared in a Civil
Service examination and created so much interest that the New
Statesman persuaded Phillips to set similar problems in place of their
crossword and bridge columns. Note the class distinctions involved in
addressing the workers by their surnames alone, and the passengers
with the prefix 'Mister'.
336. 'The driver, fireman and guard of a certain train were Brown,
Robinson and Jones, and the passengers included Mr Brown, Mr
Robinson and Mr Jones. Mr Robinson lived in Leeds; the guard lived
midway between Leeds and London. Mr Jones' income is £400 2s Id,
and the guard's income is exactly one-third of his nearest passenger
neighbour. The guard's namesake lives in London. Brown beats the
fireman at billiards. What is the name of the engine-driver?'
337. Stritebatt '''Had a good season?" I enquired of my cricketing
friend, Stritebatt.
'''Fairish. I finished with an average of exactly 30."
'''For how many innings?"
'''I forget. But we seldom had more than one in an afternoon, you
know. And I've only been able to play on Saturdays."
'''Thirty is pretty good," said I.
'''Not bad," said Stritebatt. "But I was Not Out several times, you
know. My friend Smith worked out that if I'd scored another dozen,
in each of my Not Out innings, my average would have been 35."
'" Any good scores?" asked I.
'''Nothmg special. Two blobs. Otherwise, my lowest score was 17.
By the way, I had no two scores the same, apart from the blobs, I
mean. And all my best scores were Not Out ones."
'What was Stritebatt's hIghest score?'
338. The Lodger's Bacon 'At Aspidistra House it is the landlady's
custom to put the breakfast bacon on a dish before the fire, so the
114 Penguin Book of Curious and Interesting Puzzles
lodgers may help themselves as they come down in the morning. On
this particular morning, all had shared alike, there would have been a
whole number of rashers each, but Smith, who came down last,
found only half a rasher left for him. Jones is very regular, and
always takes one rasher; Robinson takes his fair share of what
remains when he comes down; Brown is greedy, and takes his fair
share and then half a rasher extra; Evans likes three rashers, but,
being superstitious, always leaves one at least for those who follow
him.
'How many rashers did Evans take?'
339. Falsehoods 'Messrs Draper, Grocer, Baker and Hatter are (ap-
propriately enough) a draper, grocer, baker and hatter. But none of
them is the namesake of his own vocation.
'When 1 tried to find out who is who, four statements were made
to me: (1) "Mr Draper is the hatter." (2) "Mr Grocer is the draper."
(3) "Mr Baker is not the hatter." (4) "Mr Hatter is not the baker."
But clearly there was something wrong here, since Mr Baker is not
the baker.
'I subsequently discovered that three of the four statements made
to me are untrue.
'Who is the grocer?'
340. 'Alice was Disconcerted' '''What's 19 times 19?" asked the
Red Queen.
'''361,'' said Alice.
'''Wrong,'' said the Red Queen. "The answer's 519."
'Alice was disconcerted. They had forgotten to tell her that, just to
make things difficult, each digit in this multiplication represented a
different one.
'What number does "19" represent?'
341. Moulting Feathers 'Last summer I spent a week or so in the
little-known village of Moulting Feathers, in Dumpshire. Its social
centre is the local Bird Fanciers' Club.
'The club has seven members. Each is the owner of one bird. And
each owner is, strange to say, the namesake of the bird owned by one
of the others.
'Three of the fanciers have birds which are darker than their
owners' feathered namesakes.
'I stayed with the human namesake of Mr Crow's bird, from
whose wife I collected most of the village gossip. Incidentally, only
The Puzzles 115
two of the fanciers - Mr Dove and Mr Canary - are bachelors.
'Mr Gull's wife's sister's husband is the owner of the raven - the
most popular of the seven birds. The crow, on the other hand, is
much disliked; "I can't abide him," said his owner's fiancee.
'Mr Raven's bird's human namesake is the owner of the canary,
while the parrot's owner's feathered namesake is owned by the
human namesake of Mr Crow's bird.
'Who owns the starling?'
342. Windows 'Sir Draftover Grabbe, Chancellor of the Exchequer,
conceived the unoriginal plan of imposing a tax on windows. A
window having 12 square feet of glass paid a tax of £2 3s; a 24-
square-foot window paid £3 Is; a 48-square-foot window paid £4
17s.
'What do you suppose was the basis of the tax?'
Looking-glass Zoo This puzzle was composed by the famous astro-
physicist and philosopher Sir Arthur Eddington, and originally pub-
lished by Hubert Phillips in Question Time:
343. 'I took some nephews and nieces to the Zoo, and we halted at a
cage marked
Tovus Slithius, male and female.
Borogovus Mimsius, male and female.
Rathus Momus, male and female.
Jabberwockius Vulgaris, male and female.
The eight animals were asleep in a row, and the children began to
guess which was which. "That one at the end is Mr Tove." "No, no!
It's Mrs Jabberwock," and so on. I suggested that they should each
write down the names in order from left to right, and offered a prize
to the one who got most names right.
'As the four species were easily distinguishable, no mistake would
arise in pairing the animals; naturally a child who identifi"ed one
animal as Mr Tove identified the other animal of the same species as
Mrs Tove.
'The keeper, who consented to judge the lists, scrutinized them
carefully. "Here's a queer thing. I take two of the lists, say, John's
and Mary's. The animal which John supposes to be the animal which
Mary supposes to be Mr Tove is the animal which Mary supposes to
be the animal which John supposes to be Mrs Tove. It is just the
same for every pair of lists, and for all four species.
116 Penguin Book of Curious and Interesting Puzzles
"'Curiouser and curiouser! Each boy supposes Mr Tove to be the
animal which he supposes to be Mr Tove; but each girl supposes Mr
Tove to be the animal which she supposes to be Mrs Tove. And
similarly for the other animals. I mean, for instance, that the animal
Mary calls Mr Tove is really Mrs Rathe, but the animal she calls Mrs
Rathe is really Mrs Tove."
'''It seems a little involved," I said, "but I suppose it is a remarkable
coincidence. "
"'Very remarkable," replied Mr Dodgson (whom I had supposed
to be the keeper) "and it could not have happened if you had brought
any more children."
'How many nephews and nieces were there? Was the winner a boy
or a girl? And how many names did the winner get right?'
344. Wheels around Wheels How many times does a coin rotate
in rolling completely about another coin, of the same size, without
slipping?
345. Covering a Chessboard If two squares are removed from a
chessboard, one from each end of one of the long diagonals, can the
squares that remain be covered by thirty-one dominoes, each large
enough to exactly cover a pair of adjacent squares?
346. Too Many Girls In a far off land where warfare had raged for
many years, the number of men was too few for the number of
women who wished to marry them. While nothing could be done
immediately about this sorry state of affairs, the King was determined
that in future there should be more boys born and fewer girls, in
anticipation of the ravages of war.
With this aim in mind, he decreed that every woman should cease
to bear children as soon as she gave birth to her first daughter,
reasoning that while there would be some families which would have
only one daughter, or even one son and one daughter, there would be
many with seyeral sons followed by a single daughter, producing an
overall surplus of sons.
Where did his ingenious scheme go wrong?
347. Forty Unfaithful Wives 'The great Sultan was very much wor-
ried about the large number of unfaithful wives among the population
of his capital city. There were forty women who were openly deceiving
their husbands, but, as often happens, although all these cases were a
matter of common knowledge, the husbands in question were ignorant
The Puzzles 117
of their wives' behaviour. In order to punish the wretched women,
the sultan issued a proclamation which permitted the husbands of
unfaithful wives to kill them, provided, however, that they were quite
sure of the infidelity. The proclamation did not mention either the
number or the names of the wives known to be unfaithful; it merely
stated that such cases were known in the city and suggested that the
husbands do something about it. However, to the great surprise of
the entire legislative body and the city police, no wife killings were
reported on the day of the proclamation, or on the days that followed.
In fact, an entire month passed without any result, and it seemed the
deceived husbands just did not care to save their honour.
"'0 Great Sultan," said the vizier, "shouldn't we announce the
names of the forty unfaithful wives, if the husbands are too lazy to
pursue the cases themselves?"
'''No,'' said the Sultan. "Let us wait. My people may be lazy, but
they are certainly very intelligent and wise. I am sure action will be
taken very soon."
'And, indeed, on the fortieth day after the proclamation, action
suddenly broke out. That single night forty women were killed, and a
quick check revealed that they were the forty who were known to
have been deceiving their husbands.
'''I do not understand it," exclaimed the vizier. "Why did these
forty wronged husbands wait such a long time to take action, and
why did they all finally take it on the same day?'"
•
348. What Moves Backwards When the express from Bristol to
London is thundering towards London, some parts of the train are, at
one moment, moving towards Bristol. Which parts?
349. The Backwards Bicycle A bicycle is supported vertically, but is
free to move forwards and backwards when the handlebars are pushed.
One pedal is at its lowest point and the other is at its highest point.
If a string is attached to the lower pedal, and pulled backwards,
will the bicycle move forwards or backwards?
350. The Heads of Hair There are at least 50 million people living
in the United Kingdom, and no human being has more than a million
hairs on their head. What is the least number of inhabitants of the
United Kingdom who must have, according to the information,
exactly the same number of hairs on their head?
118 Penguin Book of Curious and Interesting Puzzles
351. Quickies (a) 'Have you ever seen anyone running along the pave-
ment and placing their feet on the ground in this order: right foot,
right foot, left foot, left foot, right, right, left, left ... ?'
(b) 'By suitably placing a six-inch square over a triangle I can cover up
to three-quarters of the triangle. By suitably placing the triangle over
the square, I can cover up to one-half of the square. What is the area
of the triangle?'
(c) 'When is it polite to overtake, or pass, on the inside only?'
(d) '''Don't forget you owe me five pence!" said Fred.
'''What!'' replied Tom, "Five pence isn't worth bothering about."
'''All right then," said Fred, "you can give me ten pence."
'What is the logic behind Fred's reply?'
352. An Amazing Escape 'Archaeologists, more than most scientists,
destroy cherished myths with every discovery they make. When they
claim, however, that the Labyrinth which trapped Theseus was merely
the rooms of a palace with which he was unfamiliar, they are going
The Puzzles 119
too far. They fail to appreciate that when you are being hotly
pursued by a Minotaur, you must take every opportunity to turn left
or right to escape the beast, however quickly you might otherwise
escape in a straight line. Naturally Theseus, who entered by the south
entrance to the palace, wanted to get to his beloved Ariadne, who
was waiting just outside the north entrance, as quickly as possible. What
was his shortest route If he was to evade the Minotaur?'
353. Speedy Gonzales The other day, travelling by London under-
ground, I dashed on to the platform just as my train was moving out.
I caught the next one, and left the exit of my destination station at
exactly the same time as I would have done had I not missed the first
train. Both trains travelled at the same speed, no acts of God were
involved, and I didn't have to rush to make up for lost time.
Explanation please?'
354. An Intimate Affair 'At an intimate little soiree given by Lady
What's-her-name the other evening, each man danced with exactly
three women and each woman with exactly three men. What is more,
each pair of men had exactly two dancing partners in common. An
admirable arrangement which pleased Lady What's-her-name no end
and also gives the reader enough information to discover exactly how
intimate that soiree was. How many people attended?'
355. Rice Division 'Mr and Mrs Lo Hun were poor peasant farmers,
so when Mrs Lo Hun accidentally smashed the measuring bowl
which she used for measuring out the rice, she was very upset.
Fortunately, her husband was skilled in the traditional art of sword-
fighting, and she brightened considerably when he took a strong
cardboard box of rectangular shape and, with the minimum necessary
number of clean plane sword cuts, produced a substitute for her bowl
which actually measured out one, two, three or four measures of rice,
according to her choice.
'How many cuts did her husband make, and what shape was the
final article?'
356. A Mon-ster Puzzle The dlustratlon overleaf shows a mon, a
Japanese family crest. Janet wanted to show it in her project on
Japanese history, and she was just about to cut two equal squares of
gummed paper, one black and one white, into quarters and stick
them down so that they overlapped in sequence, when it struck her
that eight pieces might not be necessary. Indeed, they were not, and
120 Penguin Book of Curious and Interesting Puzzles
eventually she managed to make the mon by using fewer pieces. How
many separate pieces did she use?
357. Knot these Cubes What is the shortest knot that can be tied in
three dimensions using only face connected cubes? All the cubes are
the same size, the knot must be continuous with no loose ends, and
the cubes must be connected by complete faces.
358. These twelve matches form one square and four triangles. How
can half of them be moved to form one triangle and three squares?
The Puzzles 121
359. Two Squares in One Each of these four pieces has two right-
angles, so it is hardly surprising that they can be fitted together to
form a square in more than one way. In how many, precisely?
360. Half-hearted Betting Major Watson, despite being short of the
readies, was feeling in a good mood, so he decided to gamble with his
friend Butterworth on the toss of a coin. Starting with just £1, he bet
six times, each time wagering half the money he had at the time.
If he won three times and lost three times, how much did he win or
lose in the end?
361. Chopsticks 'A firewood merchant had a number of blocks to
chop up for firewood. He chopped each block into eleven sticks.
Assuming that he chopped at the average rate of forty-five strokes per
minute, how many blocks would he chop up in twenty-two minutes?'
362. The Convivial Visitor '''There are only four pubs in this
village," the visitor was informed, "one in each street. The village's
four streets meet at the crossroads at right-angles. This street is the
High Street.
'''To reach the Blue Boar from the Griffin you must turn left. To
reach the Dragon from the Red Lion you have to turn right."
'The visitor entered three of the pubs; he arrived at the crossroads
three times during this pilgrimage, turning left the first time, going
straight across the second, and turning right the third time. He spent
the night at the Blue Boar.
'Which pub stands in the High Street?'
363. Siding by Siding These five locomotives have to be driven into
their respective sheds, marked with their number. Moving an engine
into a shed and out again, without moving any other engine in the
122 Penguin Book of Curious and Interesting Puzzles
meantime, counts as one move, so for example moving A into e, out
to B and into d, would count as just one move.
The puzzle is to do this as efficiently as possible.
364. Triangles within Triangles The sides of a triangle have each
been divided into quarters, and each vertex joined to one of the
points of division, as shown in the figure.
What is the area of the triangle in the centre?
365. A Unique Number What is the unique whole number whose
square and cube between them use up each of the digits 0 to 9, once
each?
366. The Picnic Ham 'Three neighbours gave $4 each and bought a
ham (without skin, fat, and bones). One of them divided it Into three
parts asserting that the weights were equal. The second neighbour
The Puzzles 123
declared that she trusted only the balance of the shop at the corner.
There, it appeared that the parts, supposed to be equal, corresponded
to the monetary values of $3, $4 and $5, respectively. The third
partner decIded to weigh the ham on her home balance, which gave a
stilI different result. This led to a quarrel, because the first woman
kept insIsting on the equality of her division, the second one recog-
nized only the balance of the shop, and the third only her own
balance. In what way IS it possible to settle this dispute and to divide
these pieces (without cutting them anew) in such a way that each
woman would have to admit that she had got at least $4 worth of
ham if computed according to the balance which she trusted?'
367. Giants and Midgets Unlike the Household Guards, Major
Mason's Mercenaries were a right shower, ranging from tall and
skinny, to short and fubsy, not forgetting the enormous Corporal Gut
and the minuscule Private Git.
One afternoon, Major Mason decided to divide them into three
separate groups, the large, the middling and the small, who could
parade separately without provoking the jeers of the locals.
He first assembled them in a rectangular array, and instructed the
tallest man In each row to step out. From these he chose the shortest
of the tall and announced that he and all those taller than him, would
form the first new platoon.
The men returned to their initial positions, and the smallest man in
each column was ordered to step out, and Major Mason picked out
the tallest of the shortest, declaring that he and all the soldiers shorter
than he would form the third platoon. The remainder would form the
second platoon.
To the Major's rage and disgust, not only were there apparently no
men at all in the second platoon, but Private Ponce claimed that he
was in both the first and third platoons.
The Major naturally flogged every man jack of them severely, and
declared that they would do it again, and properly this time. What
was the result?
368. Near Neighbours Trevor the travel agent has a map of Europe
on which every major town is joined to the town nearest to it. The
distances between towns are always different, when measured suffi-
cientlyaccurately.
What is the largest number of other towns to which anyone town
can be connected?
124 Penguin Book of Curious and Interesting Puzzles
369. Guarding the Gallery The new art gallery has twenty walls,
each wall being at right-angles to its adjoining walls. Without know-
ing the precise design of the gallery, what is the least number ·of
guards that will guarantee that all the walls can be kept under
observation all the time?
370. Batty Batting Frank had an excellent first half of the season,
averaging comfortably more runs per innings than Paul. Moreover he
had started the second half of the season very well, and he looked
forward to once again picking up the club trophy for best overall
batting average.
At the end of the second half of the season, Frank had indeed once
again beaten Paul's average, yet for the whole season, to Frank's
disgust, Paul was ahead, and took the trophy. How was this possible?
371. The Bouncing Billiard Ball A mathematical billiard table is in
the form of a rectangle with integral sides, and just four pockets, one
in each corner. A ball shoots out of one pocket at angles of 45° to the
sides. Will it bounce round the table for ever, or end up in one of the
other pockets?
372. Back to the Start A billiard ball is struck without side so that
it strikes all four cushions and returns to its starting position.
In what direction is it struck, and how far does it travel?
373. Lies, Almost All Lies Here are ten numbered statements. How
many of them are true?
1 Exactly one of these statements is false.
2 Exactly two of these statements are false.
3 Exactly three of these statements are false.
4 Exactly four of these statements are false.
5 Exactly five of these statements are false.
6 Exactly six of these statements are false.
7 Exactly seven of these statements are false.
8 Exactly eight of these statements are false.
9 Exactly nine of these statements are false.
10 Exactly ten of these statements are false.
374. Bookworm A bookworm, feeling very hungry, is delighted to
come across the three volumes of Dr Johnson's great Dictionary of
the English Language standing on a shelf. Starting from the front
The Puzzles 125
cover of the first volume, it bores its way through to the back of the
back cover of the third volume.
If the front and back covers of each volume are t cm thick and the
pages of each volume are 7 cm thick, how far does the bookworm
bore?
375. OH-HO How many moves are required to transform this H
into the 0, if a move consists of sliding one coin to touch two others,
without moving any of the other coins?
When you have changed the H to 0, how many moves does it take
to get back from 0 to H?
376. Inverted Triangle This triangle contains ten coins. What is the
smallest number that must be moved to make the triangle point down-
wards?
377. Paradoxical Dice Alan, Barry and Chris were playing at dice,
using three fair dice which they had each marked with their own
special numbers. Alan consistently beat Barry, and Barry's dice consist-
ently beat Chris's. What was surprising was that Chris's dice neverthe-
less consistently beat Alan's.
How was this possible?
126 Penguin Book of Curious and Interesting Puzzles
378. The Obedient Ray Two mirrors are joined at a fixed angle at
0, and a ray of light is shone into the angle between them, parallel to
one of the mirrors. It bounces a number of times, strikes the lower
mirror at right-angles at X, and then re-emerges along its original path.
O ~ - - - - - - ~ - - ~ - - - - - - - - - - - - - - - - - - -
X
What is the distance between the original ray and the lower mirror,
to which it is parallel?
379. The Balloon 'Mr Tabako's little boy sits in the back seat of a
closed motor-car, holding a balloon on a string. All the windows of
the car are closed tight. The balloon is full of coal gas and is tethered
by a string, which prevents it from touching the roof of the car.
'The car turns left at a roundabout. Does the balloon swing left,
swing right, stay upright, or do something else? And why?'
380. Which Contains the Beer? A grocer has six barrels of different
sizes, containing 15, 16, 18, 19,20 and 31 litres. Five barrels are filled
with wine and only one is filled with beer.
The first customer bought two barrels of wine, and a second
customer also bought wine, but twice as much as the first. Which is
the beer barrel?
381. Blackbirds
'Twice four and twenty blackbirds
Were sitting in the rain.
Jill shot and killed a seventh part.
How many did remain?'
382. The Two Girlfriends John is equally devoted to his two girl-
friends, one of whom lives uptown and the other downtown. He
therefore decides to catch the first bus to arrive, whichever direction
it is going in. Since all the buses run at equal intervals, and his own
The Puzzles 127
times of arrival at the road are quite random, he looks forward to
visiting each girl with equal frequency, yet he soon finds out that he is
seeing one far more often than the other. Why?
383. Confounded Cancellation Mr Peebles believed in giving his
pupils responsibility, so when he had to leave the class one day he
instructed Jones Minor to go to the board and write up some simple
fraction sums which the rest of the class were to do until he returned.
On his return, he found the class rolling in the a'isles with laughter.
Looking at the board he saw that the first 'sum' was written as
1 ~ =_
~ 4
The other three sums were also fractions with numerators and
denominators below 100, and each was simplified by Jones Minor in
the same absurd - but in this case 'correct' - manner. What were
they?
384. Back to Back These cards are used in a simple psychological
test. Every card certainly has a letter on one side and a number on the
other. I make the additional claim to you that if you look at the letter
on one side of a card and see that it is a vowel, then you can be
certain that the number on the other side is even.
A 2 3 B
How many cards must you turn over to check whether my additional
claim is correct?
385. The Cigarette Ends A tramp collecting cigarette ends from the
street can make a new cigarette out of four ends. He collects in one
morning, thirty-two ends. How many cigarettes can he smoke that
day?
386. Coin Catch I have only two coins in my pocket. They add up
to 15 pence and yet one of them is not a 10 pence piece. What
denominations are they?
128 Penguin Book of Curious and Interesting Puzzles
387. Beer from a Can You are drinking beer from a can. When the
can is full, the centre of gravity of the beer and can together will be in
the centre of the can, as near as makes no difference. As you start to
drink, the centre of gravity falls, but by the time the can is empty it is
back to the centre of the now-empty can.
At what point did the centre of gravity reach its minimum
position?
388. Long-playing Poser How many grooves are there on a standard
long-playing record?
389. The Lily in the Pond A water lily doubles in size, that is, in the
area of the leaf lying on the surface of the pond, every 24 hours. If it
takes 30 days to cover the pond completely, after how many days did
it cover exactly one half of the pond?
390. A Shaking Result At a recent conference, most of the delegates,
but not all, shook hands with most of the other delegates on arrival,
and again on leaving. Why was the number of delegates who shook
hands an odd number of times necessarily even?
391. The Hotel Reception A group of travellers, seven in number,
arrived at a hotel and asked to be put up for the night. The manager
actually only had six rooms available, but he promised to do what he
could.
First he put the first man in the first room and asked another man
to wait there for a few minutes. He then placed the third man in the
second room, the fourth in the third room, and the fifth in the fourth
room. Finally he placed the sixth man in the fifth room and went
back for the seventh man, who he placed in the sixth room. OK?
392. An Irrational Number 'Show, by a simple example, that an
irrational number raised to an irrational power need not be irra-
tional.'
393. Horseshoe Dissection How can a horseshoe be cut into SIX
separate pieces with just two cuts?
394. A Price Poser
'How much will one cost?'
'Thirty pence.'
'How much will fifteen cost?'
'Sixty pence.'
'Thank you, I'll take one hundred and sixteen.'
'That will be ninety pence, Madam.'
Explain, please.
The Puzzles 129
395. Hula Hoop 'Consider a vertical girl whose waist is circular,
not smooth, and temporarily at rest. Around her waist rotates a hula
hoop of twice its diameter. Show that after one revolution of the
hoop, the point originally in contact with the girl has travelled a
distance equal to the perimeter of a square circumscribing the girl's
waist.'
396. Up and Down John Smith leaves home every morning, from
his flat at the top of a tower block, and takes the lift to the ground
floor, walks to the bus stop and catches the bus.
On the way home, however, he gets off the bus, walks to the tower
block entrance, takes the lift to the seventh floor and then walks the
rest of the way. Why? It may help you to know that he is extremely
healthy, and is not in need of exercise.
130 Penguin Book of Curious and Interesting Puzzles
397. Three into Two You have a frying pan which will take only
two slices of bread at a time, and you wish to fry three slices, each on
both sides. Since each slice takes 20 seconds for each side, you can
certainly fry them all in 80 seconds, by doing two pieces together and
then the third.
But can you fry them more efficiently?
398. The Jigsaw Puzzle 'In assembling a jigsaw puzzle, let us call
the fitting together of two pieces a "move", independently of whether
the pieces consist of single pieces or of blocks of pieces already
assembled. What procedures will minimize the number of moves
required to solve an n-piece puzzle? What is the minimum number of
moves needed?'
399. The Ladder and the Box A ladder, 4 metres long, is leaning
against a wall in such a way that it just touches a box, 1 metre by 1
metre, as in the figure. How high is the top of the ladder above the
floor?
The Puzzles 131
400. The Crossed Ladders Two ladders, 20 and 30 feet long, lean
across a passageway. They cross at a point 8 feet above the floor.
How wide is the passage?
401. Odd Corners The regular tetrahedron and the regular dodeca-
hedron both have vertices at which an odd number of edges meet, but
they have an even number of such vertices.
Is it possible for a polyhedron to have an odd number of vertices at
which an odd number of edges meet?
402. Where does Mr Jones Live? Mr Jones has moved to a new house
in a rather long street, and has noticed that the sum of the numbers up
to his own house, but excluding it, equals the sum of the numbers of his
house to the end house in the road. If the houses are numbered
consecutively, starting from 1, what number does Mr Jones live at?
403. The Knockout Tournament If the number of players entered
for a knockout tournament is a power of 2, for example 8, 16 or 32,
then it is easy to arrange the pairings and it is obVIOUS how many
matches will take place in each round.
What happens If there is a different number of entrants? In
particular, how many matches will have to he played if thirty-seven
players enter a knockout?
132 Penguin Book of Curious and Interesting Puzzles
404. The Breakfast Egg Mr Oval started every day with an egg,
lightly boiled, with a slice of toast, yet he never bought an egg,
neither borrowed nor stole his eggs and did not keep chickens. Please
explain!
405. A Riddle Two legs sat on three legs when along came four legs
and stole the one leg, whereupon two legs picked up three legs and
threw it at four legs, and got his one back. Explain, please.
406. The Bottle and Cork A bottle and its cork cost 21 pence and
the bottle costs 20 pence more than the cork. What is the cost of
each?
407. The Mixed-up Labels You are given three boxes containing,
respectively, chocolate drops, aniseed balls, and a mixture. Unfortu-
nately, every jar has been wrongly labelled with the label that ought
to have gone on one of the other jars.
What is the least you need do to discover which jar is which and
restore the labels to their correct jars?
408. Knotted or not Knotted? Is this piece of rope genuinely knotted,
or just in a tangle? In other words, what will happen if you pull the two
ends apart? Wil\.it tighten into a knot, or stretch into a straight line?
The Puzzles 133
409. The Circle and Saucers 'Our table top is circular and its
diameter is fifteen times the diameter of our saucers, which are also
circular. How many saucers can be placed on the table top so that
they overlap neither each other nor the edge of the table?'
410. Balls in a Box What is the size of the smallest cubical box
which will just contain four balls, each ten inches in diameter?
411. The Problem of the Calissons Calissons are a French sweet, in
the shape of two equilateral triangles edge to edge, which come
packed in a hexagonal box. As you can see from the figure, the
calissons can be pointing in any of three directions.
Your puzzle is to explain why the number of caltssons pointing in
each direction, In a fully packed box, must always be equal.
412. The Long Shot A hunter travelling by train to the forest
carries with him his gun, which is 2.3 metres long. Unfortunately, the
baggage regulations of the train company forbid any object more
than 2 metres long. How does the hunter get round this rule?
413. Cigarette Extras A manufacturer produces boxes of cigarettes.
Each box contains 160 cigarettes arranged In eight rows of twenty.
134 Penguin Book of Curious and Interesting Puzzles
Assuming that the cigarettes completely fill the box, it is neverthe-
less possible to get more cigarettes into the box. How can this be
done, and how many extra can be accommodated?
414. A Moving Poser Place three coins in a row, like this, so that
each touches the next:
A 8 c
The puzzle is to move coin A so that it is between coins Band C,
without touching either B or C.
415. High Stakes 'Mike sat down and started shuffling the cards.
"What stakes?" he asked.
'''Let's make it a gamble," Steve replied, putting a few bills and
some coins in the table. "The first game, the loser pays 1 cent, the
second 2 cents, and sopn. Double up each time."
"'Okay," laughed Mike, checking his cash, "I've got only $6.01
and I'm not playing more than ten games anyway."
'So they played, and game followed game until at last Mike stood
up. "That's my last cent I've just paid you," he declared, "but I'll
have my revenge next week."
'How many games had they played, and which did Mike win?'
416. Some are Less Equal than Others 'A pencil, eraser and note-
book together costs $1. A notebook costs more than two pencils, and
three pencils cost more than four erasers. If three erasers cost more
than a notebook, how much does each cost?'
417. Pandigital Probability What IS the probability that a ten-digit
number, that is, a number chosen at random between 1,000,000,000
and 9,999,999,999 inclusive, will have ten different digits?
418. The Broken Stick (1) A stick is broken into three pieces. What
is the probability that they will form a triangle?
To make what we have in mmd a little clearer, let's say that two
points are chosen at random on the stick, each choice being independ-
ent of the other, and the stick is broken at those points.
The Puzzles 135
419. The Broken Stick (2) A stick IS broken into two pieces, at
random. What IS the average length of the shorter piece?
420 A Striking Clock When a grandfather clock strikes 6 o'clock,
there are 15 seconds between the first and last strokes. How many
seconds elapse between the first and last strokes when it strikes mid-
mght?
421. Buried Treasure A treasure is bUrled somewhere along a
straight road on which are four towns, the distances between them, in
sequence, being 5, 8 and 11 miles.
A map gives the following instructions for finding' the treasure.
Unfortunately, as indicated by the words in quotation marks, the
actual names of the towns have become illegible with age. Despite
this difficulty, the site of the treasure can be located. How?
Start at town 'squiggle' and go half of the way to 'splodge'.
Then go one third of the way to 'can't read it' and finally travel
one quarter of the way to 'd1egible'.
422. Multiple Ages A man and his grandson have the same birthday.
For six consecutive birthdays the man is an integral number of times
as old as his grandson. How old is each at the sixth of these birth-
days?
423. Grandfather and Grandson 'In 1932 I was as old as the last
two digits of my birth year. When I mentioned this interesting
coinCidence to my grandfather, he surprised me by saying that the
same applied to him too. I thought that impossible .. .'
What were their ages?
424. Diagonals of a Cube This figure shows two face diagonals of a
cube. What is the angle between them?
136 Penguin Book of Curious and Interesting Puzzles
425. Equal Areas How many regions of equal area can you see in
this figure?
426. A Great Day for the Race 'Fred Bretts noticed that there were
nine runners in the big race and asked his bookie what odds he was
offering.
'''3-1 on Bonnie Lass, 4-1 on Golden Stirrup, 7-1 on Two's a
Crowd, 9--1 on Greek Hero and 39-1 the field," he replied.
'Fred thought for a few moments and then astounded the bookie
by placing a bet on each of the nine horses, all to win. No each-way
nonsense for fearless Fred. And all on credit, of course.
'''You might as well give me my winnings now," said Fred.
'''The race hasn't been run yet, Sir," smiled the bookie.
'''That doesn't matter," said Fred. "When it has, you'll owe me
£200. "
'And he was right.
'How much did he stake on each horse?'
427. Find the Centre How can the centre of a circle be found,
accurately, by the use of a set-square only?
428. The Trisected Angle 'A confirmed angle-watcher one afternoon
ruled two lines on a clock face to mark the angle then formed by the
two hands. Some time later he noticed that the two hands exactly
trisected the angle he had marked. In how short an interval could this
have happened? And how soon after 3 o'clock could he have ruled his
lines in order to observe the trisection in this short a time?'
The Puzzles 137
429. Popsicle Polygons The figure shows how five iced-lolly sticks,
called popsicle sticks in the United States, can be used to make a
triangle that can be picked up and handed round without falling
apart.
How many sticks are needed to make a regular hexagon?
430. Touching Three What is the smallest number of pennies that
must be placed on a table for each penny to touch exactly three
others, .f every coin is flat on the table?
431. Eight Heads and Eight Tails Lay down sixteen coins, heads
and tails alternately as shown. The problem is to rearrange the coins
so that those in each vertical column are alike. Two coins only, may
be touched.
138 Penguin Book of Curious and Interesting Puzzles
432. FaIling on Edge 'How thick should a coin have to be to have a
1 in 3 chance of landing on edge?'
433. A Simple Angle How can an angle of 30° be constructed using
only an unmarked ruler?
434. Bisecting the Segment You have an unmarked ruler, whose
opposite edges are parallel. How can you bisect a given line segment,
which is shorter than your ruler?
435. A Moving Problem Jack and Jill are moving to a new flat and
their grand piano presents a potential problem. Fortunately, it will
just pass round the corridor without being tipped on its end or being
disassembled.
Given that its area, on looking down on it from above, is the
largest possible which can be passed round the corner, what are the
proportions of its length to its width?
436. Square and Add The number 3025 has the curious property
that if you split it into two parts, add the two parts together and
square the result, the original number is recovered:
30 + 25 = 55 and 5SZ = 3025.
What is the only other number, consisting of four different digits,
with this property?
The Puzzles 139
437. Striking a Balance 'Having a parcel to send by post, and being
uncertain whether it was under 2 pounds (or 32 ounces) in weight,
when it would go for sixpence, or heavier than that, which would
mean ninepence, I borrowed some rather primitive scales from my
landlady.
'The first weighing gave 28t ounces, well within the sixpenny
range, but, being rather doubtful of the balance, which looked as if
one arm was longer than the other, I tried weighing the parcel in the
other scale-pan. This gave the weight as 36 ounces, thoroughly
justifying my suspicions.
'How should I stamp my parcel?'
438. Dud Coins by the Boxful Mr Jones has plenty of coins, ten
boxes of them in fact, but unfortunately one box contains duds which
are all 2 gm short in weight. Even more unfortunately, he has forgot-
ten which box contains the duds. If all the other boxes contain good
coins, weighing 40 gm each, how many weighings on a weighing
machine are necessary to decide which box has the duds?
439. The Problem of Twelve Coins 'Among twelve coins there is
one at the most which has a false weight. With three weighings on an
equal-arm balance, but with no use of weights, show how to establish
whether there is a false coin, or not; and if so, which it is, and
whether it is too light or too heavy.'
440. A Sound Bet? 'I will bet you one pound,' said Fred, 'that if you
give me two pounds, I will give you three pounds in return.'
'Done,' replied Jack. Was he?
441. Sealed Bids 'Red and Black each stakes a 5 pence piece. Now
each competes for this pool by writing down a sealed bid. When the
bids are simultaneously revealed, the high bidder wins the stakes but
pays the low bidder the amount of his low bid. If the bids are equal,
Red and Black split the stakes.
'How much do you bid, Red?'
442. Sharing the Sandwiches Jones and Smith were sharing a jour-
ney, and when they felt hungry they prepared to share their sand-
wiches, of which Jones had brought five and Smith had brought
three.
However, seeing a stranger, who turned out to be Mr Watson,
eyeing their sandwiches, they offered to share them with him. Watson
140 Penguin Book of Curious and Interesting Puzzles
accepted, and they shared the sandwiches equally, after which Watson
insisted on contributing £2 to the coSt of his lunch.
Jones immediately suggested that they split the money in proportion
to their contributions, Jones taking five parts, or £1.25, and Smith
taking three parts, or 75p. But Smith objected, insisting that this
would not be just. Who was right, and how much did each receive?
443. Pulling a Pint ' A stranger walked into a public bar, put ten-
pence on the counter and asked for half a pint of beer. The barmaid
asked whether he would like Flowers or I.P.A. The stranger asked for
Flowers.
'Another complete stranger entered the bar, put tenpence on the
counter and asked for half a pint of beer. Upon which the barmaid
immediately pulled half of Flowers. How did she know what the
second man, who was a stranger to her, wanted?'
444. A Square Chessboard How many squares are there on an 8 x
8 chessboard?
445. Mid-point with Compass Only You are given two points which
may be thought of as the ends of a line segment, except that the line
isn't there. How can you find the mid-point of the imaginary segment
using only a pair of compasses? No ruler, no straight-edge are
allowed, and no folding to get a straight line by stealth.
446. Tricky Tumblers Here are six tumblers, three full and three
empty, arranged in a row.
UD DUU
What is the smallest number of moves needed to leave the tumblers
alternately full and empty? Every time a tumbler is picked up, that
counts as a move.
447. Turning Tails There are eight ways to arrange three coins in a
row, each coin showing either head or tail. Starting with three heads
showing, and changing only one coin at a time, can you in just seven
The Puzzles 141
turns go through the entire sequence, ending up with three tails
uppermost?
448. Little Pigley Farm, 1935 The first crossword appeared on 21
December 1913, in The New York World, but crosswords did not
take off until 1924 when Simon and Schuster published a book of
fifty puzzles. Crossword mania erupted, everyone jumped on to the
bandwagon, and by the end of the first year 350,000 crossword books
had been sold.
Naturally, cross-numbers were soon to follow. This puzzle is from
The Strand Problems Book by W. T. Williams, who composed
puzzles for John O'London's Weekly, and G. H. Savage, who pub-
lished in The Strand Magazine.
Note: One of the 'across' numbers is the same as one of the 'downs'.
This is the only case of identity, though one number in the puzzle
(relating to something quite different) happens to be the area in roods
of the rectangular field' known as Dog's Mead. Equipped with this
information and the homely items that follow, the reader is invited to
discover that jealously guarded secret, the age of Mrs Grooby, Farmer
Dunk's mother-in-law.
Readers may like to know that 1 acre was 4840 square yards, and 1
rood was one quarter of an acre. Also there were 20 shillings in £1 ster-
ling.
Across
I. Area of D o g ' ~ Mead 111 square
yards.
Down
1. Value 111 shillings per acre of
Dog's Mead.
142 Penguin Book of Curious and Interesting Puzzles
5. Age of Farmer Dunk's 2. The square of Mrs Grooby's
daughter, Martha. age.
6. Difference in yards of length 3. Age of Mary, Farmer's
and breadth of Dog's Mead. youngest.
7. Number of roods in Dog's 4. Value of Dog's Mead in
Mead x 9 down. pounds sterling.
8. Date (AD) when Little Pigley 6. Age of Farmer's firstborn,
came into the occupation of Ted, who will be twice as old
the Dunk family. as Mary next year.
10. Farmer Dunk's age. 7. Square of number of yards in
11. The year when Mary was breadth of Dog's Mead.
born. 8. Number of minutes Farmer
14. Perimeter in yards of Dog's takes to walk 11 times round
Mead. Dog's Mead.
15. The cube of Farmer's walking 9. See 10 down.
speed in miles per hour. 10. 10 ac. x 9 down.
16. 15 ac. minus 9 down. 12. One more than sum of digits
in column 2.
13. Length of tenure (in years) of
Little Pigley by the Dunks.
449. Fours into Nine 'This is the grid for a children's crossword, in
which no word of more than four letters will be used. Apart from this
restriction, the grid will obey the usual rule that the black squares do
not separate any part of the puzzle completely from the remainder.
'What is the smallest number of black squares that must be filled in
order to satisfy these conditions?'
450. A Common Libel
EVE
The Puzzles 143
- = .TALKTALKTALKTALK ...
DID
This represents a common fraction written as a repeating decimal.
What is the fraction?
451. 4 for Starters A number which ends in the digit 4, becomes 4
times larger when the 4 is removed from the end and placed at the
front. What is the number?
452. Twelve Up You toss a dice with the usual numbers 1 to 6 on
its faces, until the total exceeds 12. What is the most likely final total?
453. Mothers and Fathers First Only one large piece of cake re-
mained, in the shape of a triangle. 'Equal shares for all!' announced
Lilly, the tiniest.
'Agreed!' replied her mother. 'We shall all have pieces of exactly
the same shape,' and so saying she cut the triangular cake into five
pieces, all the same shape, two large and identical pieces for Father
and herself, and three smaller identical pieces for the three children.
How much more cake did Father have than Lilly?
454. Arithmetic in Pictures Anyone can see that 5' + 10' = 11' +
2', both being equal to 125. But can you demonstrate this by
geometry? Specifically by dissecting each of these figures into the
other, using of course as few pieces as possible?
455. Dollars into Cents 'When Mr Smith cashed a cheque (for less
than $100), the bank clerk accidentally mistook the number of dollars
for the number of cents, and conversely. After Mr Smith had spent 68
cents, he discovered that he had twice as much money as the cheque
had been written for. What was the amount for which the cheque had
been written?'
144 Penguin Book of Curious and Interesting Puzzles
• • •
• • •
• • •
456. Four through Nine Taking your pencil, can you cross out all
nine of these dots with four straight lines, without lifting your pencil
off the paper?
457. Sixteen Out Taking your pencil, cross out all sixteen of these
dots, in a sequence of straight strokes, without lifting your pencil
from the paper and ending up at the point where you started. How
few strokes are required?
• • • •
• • • •
• • • •
• • • •
458. The True! 'After a mutual and irreconcilable dispute among
Red, Black and Gray, the three parties have agreed to a three-way
duel. Each man is provided with a pistol and an unlimited supply of
ammunition. Instead of simultaneous volleys, a firing order is to be
established and followed until one survivor remains.
'Gray is a 100 per cent marksman, never having missed a buWs-eye
in his shooting career. Black is successful two out of three times on
the average, and you, Red, are only a 113 marksman. Recognizing the
The Puzzles 145
disparate degrees of marksmanship, the seconds have decIded that
you will be first and Black second in the firing order.
'Your pistol is loaded and cocked. At whom do you shoot?'
459. The Hurried Duellers 'Duels in the town of Discretion are rarely
fatal. There, each contestant comes at a random moment between
5 a.m. and 6 a.m. on the appointed day and leaves exactly five minutes
later, honour served, unless his opponent arrives within the time
interval and then they fight. What fraction of duels lead to violence?'
460. Matching Matches Here is a row of fifteen matches. Arrange
them in five groups of three each, by repeatedly moving one match so
that it jumps over three matches.
461. Seven Up These seven cups have to be turned the right way
up, but each move must consist of inverting three at a time. You can
choose the three from anywhere in the line, they need not for
example be adjacent, and a cup may be inverted on one move,
inverted again on the next, and so on.
f\f\f\f\f\f\f\
How many moves are necessary? How many moves would you
need if the rules specified that four cups be inverted at each turn?
462. An Objectionable Rearrangement In Class 4C there are
twenty-five desks arranged in five rows of five to form a square. On
Tuesday they had a new teacher who instructed them to each move
to a new desk, with the least fuss possible. That is, to move to the
desk either directly in front or behind, or to the right or left of their
old desk. Oh, I forgot to mention that Peaky Wilson was absent on
Tuesday and his desk remained unoccupied.
On Wednesday Peaky had returned, and once again the teacher
instructed all the students to each move to a new desk, under the
146 Penguin Book of Curious and Interesting Puzzles
same conditions as before. Peaky objected strongly, and ended up in
the Headmaster's study, accused of insolence. Why?
463. Squaring the Cube Your poser is to dissect a cube into a
square using just four pieces. No, the solution is not to cut the cube
into four identical square slices and use them as quarters of the
square - because the 'square' will then be a shallow prism.
464. Cutting the Cake Only Jane and her three closest friends are
to cut her birthday cake. If they each make one vertical cut, what is
the maximum number of pieces that they can cut?
If not all the slices have to be vertical, which is alright because the
marzipan on top makes Patrick sick anyway, how many pieces of
cake can the four of them cut?
465. A Hectic Week 'When the day after tomorrow is yesterday,
today will be as far from Sunday as today was from Sunday when the
day before yesterday was tomorrow. What day is it?'
466. Two proof readers are checking two copies of the same manu-
script. The first finds thirty errors, and the second finds only twenty-
four. When their completed proofs are compared, it turns out that
only twenty errors have been spotted by both of them.
How many errors would you suspect remain, not detected by either
of them?
467. How Many Mistakes? How many mistakes are there in this
sentence: 'This sentance contanes one misteak'?
What is the answer to the same question for this sentence: 'Their
are three misteaks in this sentence'?
468. The Professor on the Escalator 'When Professor Stamslav Slap-
enarski, the Polish mathematician, walked very slowly down the
down-moving escalator, he reached the bottom after taking fifty
steps. As an experiment, he then ran up the same escalator, one step
at a time, reaching the top after taking 125 steps.
'Assuming that the professor went up five times as fast as he went
down (that is, took five steps to everyone step before), and that he
made each trip at a constant speed, how many steps would be visible
if the escalator stopped running?'
The Puzzles 147
469. Siting a Central Depot 'The street plan of a city consists only
of straight streets intersecting at right-angles, and at an odd number
of the junctions there are kiosks. The figure gives, as an example, a
plan with ten streets and three kiosks. The occupants of the kiosks
now wish to draw their wares from a common central depot. How
should this be located so as to give a minimum total length for single
trips to the depot from each individual kiosk? The breadths of the
streets may be neglected.'
470. Nobel Prizes 'On the occasion of receiving his second Nobel
prize, Dr LinUS Pauling, the chemist, remarked that, while the chances
of any person in the world receiving his first Nobel prize were one in
several billion (the population of the world), the chances of receiving
a second Nobel prize were one in several hundred (the total number
of living people who had receIved the prize in the past) and that
therefore it was less remarkable to receive one's second prize than
one's first.'
What is the flaw in Professor Pauling's joke argument?
471. All Horses are the Same Colour Here is a proof that all horses
are the same colour. One horse is certainly the same colour as itself.
Now assume that the title statement is true of any set of N horses.
Then it follows that it is true for any set of N + 1 horses, by the
following reasoning:
Remove one horse from the set of N + 1 horses, to leave a set of N
horses who are all, by our assumption, the same colour. Next, replace
that horse and remove a different horse, to leave another set of N
horses, all the same colour. By this argument, the two horses removed
each have the same colour as the other N - 1 horses in the set.
Therefore, all N + 1 horses have the same colour.
Where is the fallacy in this argument?
148 Penguin Book of Curious and Interesting Puzzles
472. Father and Son Mr Smith and his son were involved in a
terrible accident at the factory where they worked. Mr Smith was
killed outright, and his son was rushed to the emergency unit of the
local hospital, and prepared for immediate surgery.
The surgeon on duty came into the operating theatre, saw the
patient and exclaimed, 'That's my son, I can't operate!' and sent for a
deputy.
Explain, please!
473. Father's Son Lord Elphick was showing his guest the family
portraits. Pointing to one, he remarked: 'Brothers and sisters have I
none, but that man's father is my father's son.'
Who was represented in the portniit?
474. Three Teams, New Method 'What attracts people to watch
football is goals being scored. And the authorities have been thinking
for a long time of methods whereby the scoring of more goals might
be encouraged.
'One suggestion that has been made involves a change in the way
that points are awarded. The idea is that 10 points should be
awarded for a win, 5 points for a draw, and 1 point for each goal
scored, whatever the result of the match. Therefore even if you are
losing 0-5 and have no hope of winning, a goal scored might make all
the difference between promotion and non-promotion.
'This method was tried out on a small scale and its success can be
judged from the fact that each side scored at least one goal in every
match. There were only three sides playing and eventually they are all
going to play each other once. A had scored 8 points, B had scored 14
points, and C had scored 9 points.
'Find the score in each match.'
475. Four Triangles Four identical right-angled triangles have been
added to a square, two pointing outwards, two inwards.
Why do the free vertices, marked with dots, he on a straight line?
The Puzzles 149
476. Three Digits What is the largest number that can be written
with just three digits, using no other signs or symbols at all - and
what are its last two digits?
477. The Five Couples 'My wife and I recently attended a party at
which there were four other married couples. Various handshakes
took place. No one shook hands with himself (or herself) or with his
(or her) spouse, and no one shook hands with the same person more
than once.
'After all the handshakes were over, I asked each person, including
my wife, how many hands he (or she) had shaken. To my surprise
each gave a different answer. How many hands did my wife shake?'
478. For the Love of a Good Woman Sir Pumphret and Sir Limpney
both loved the Lady Isabel and resolved to have a race, the winner to
take her hand in marriage. Knowing that Lady Isabel was opposed to
all forms of competition, they chose to have a loser's race, the one
whose horse came in last being the winner.
The first race was, predictably, a farce. They started very slowly,
then went backwards, wandered off the course, and never came
within sight of the finishing line.
The second race was quite different, both knights racing their
mounts to the finishing line. Why?
150 Penguin Book of Curious and Interesting Puzzles
479. The Egg Timer 'With a 7-minute hourglass and an II-minute
hourglass, what is the quickest way to time the boiling of an egg for
15 minutes?'
480. Kirkman's Schoolgirl Problem A schoolmistress is in the habit
of taking her girls for a daily walk. The girls are fifteen in number,
and on each walk are arranged in five rows of three, such that each
girl might have two companions. The problem is to dispose them so
that for seven consecutive days no girl will walk with any of her
school-fellows more than once.
481. Common Tangents Here are three circles and six common
tangents which meet in pairs at three points, X, Y and Z. It appears
that XYZ is a straight line.
z
When this was first shown to Professor John Edson Sweet, a famous
American engineer, Professor Sweet paused for a moment and said,
'Yes, that is perfectly self-evident.'
What was Professor Sweet's reasoning?
482. Polygon Products The figure shows how three regular dodeca-
gons can be dissected, most elegantly, into pieces which assemble into
one, larger, regular dodecagon.
Your problem is a little simpler. How can these three hexagon stars
be cut up and reassembled into one star of the same shape?
The Puzzles 151
000
483. Dodecagon into Square This IS a regular dodecagon and a
square, of equal area.
Your puzzle is to dissect each of them into six pieces that will
reassemble to form the other. Because of a hidden - well, it's not that well
hidden - feature of the two shapes, this is not as difficult as it might seem.
484. Knight's Tour On a small board, such as a 4 x 4 board, it is
not possible to start at one square and visit every square, once and
only once, making a knight's move (that is, the move of a knight in
chess) each time. Indeed, on a 4 x 4 board, wherever you start, either
four or six squares will be omitted from your tour.
What is the smallest rectangular board on which it is possible to do
a complete tour?
Can you find a board on which it is possible to make a complete
tour which has rotational symmetry, so that the tour remams un-
changed when the board is given repeated quarter-turns?
152 Penguin Book of Curious and Interesting Puzzles
485. The Bishop's Visitation What is the smallest number of moves
in which the white bishop, starting where he chooses, can visit, that
is, pass into or through, everyone of the thirty-two white squares?
486. A Handy Problem If you turn a left-handed glove inside-out,
will it be right-handed or left-handed?
487. The Magic Hexagram Twelve circles have been placed at the
vertices and intersections of this star. How can the numbers 1 to 12
be placed, one in each circle, so that the sums of the numbers in every
row, and also the sum of the six vertices, are equal?
The Puzzles 153
488. The Two Bookcases 'A room 9 by 12 feet contains two book-
cases that hold a collection of rare erotica. Bookcase A B is 8f feet
long, and bookcase CD is 4! feet long. The bookcases are positioned
so that each is centred along its wall and one inch from the wall.
9 FEET
A StFEET 8
I-
W
W
U.
C\I
c 4i FEET D
8 A
D c
'The owner's young nephews are commg for a visit. He wishes to
protect them and the books from each other by turning both book-
cases around to face the wall.' Each bookcase must end up m its
startmg position, but with Its ends reversed. The bookcases are so
heavy that the only way to move them IS to keep one end on the floor
as a Pivot while the other end IS swung in a circular arc. The
are narrow from front to back, and for purposes of the
problclll wc Idc.lltl.c them as straight Imc segments. What IS the
1llll111llUIll nUlllher of sWll1gs rCLJlI1red to reverse the two bookcases?'
154 Penguin Book of Curious and Interesting Puzzles
489. Put the Cherry in the Glass This diagram represents a cocktail
glass, composed of four matches, and a cherry.
~ , J
~ d
By moving only two matches, place the cherry into the glass.
490. The Bridge of Matches In Cambridge, over the River Cam
behind Queens' College, is a bridge which, it is claimed, was originally
designed by Sir Isaac Newton without the use of any joints or pins.
This is your chance to imitate the great man. Can you make a
bridge, using no glue or other adhesive materials or devices, using
twenty-two kitchen matches?
491. The Mystic Square This square has occult properties. For
example, careful study of the square will reveal the missing symbol
which should go into the empty cell. What is it?
Q
V W
\(J
C5
11
M C3
The Puzzles 155
492. Cannonball Quiz Charles Hutton, Professor at the Royal Mili-
tary Academy and translator of Ozanam's Mathematical Recreations,
naturally instructed his students in methods of calculating the amount
of enemy ordnance. In particular, he explained how to calculate the
number of cannonballs in a pile from the observed number along one
edge.
No doubt you can solve the puzzle without Hutton's instruction.
On this occasion you observe that the enemy have assembled their
cannonballs in one square pyramid, and you are about to raise your
telescope to count the number of cannonballs along the bottom edge,
when you see an enemy soldier walk up with one more cannonball.
The Master of Ordnance appears to remonstrate with him, and then
to take the original square pyramid apart and build a new, triangular
pyramid, using all the original balls, and the extra one.
How many cannonballs do the enemy possess?
493. Squares into Squares? Is it possible to assemble a number of
different geometrical squares, to make a larger square, leaving no
spaces? See how close you can get in assembling squares of the
following edge-lengths into a larger square: 99, 78, 77, 57, 43, 41, 34,
25,21, 16 and 9.
494. Book Words The Ruritanian National Library contains more
books than any single book on its shelves contains words. Also, no
two of its books contain the same number of words.
Can you say how many words are in one of its books?
495. Archimedes' Breath At a rough and ready estimate, what is the
probability that you are, at this very moment, breathing in at least
one molecule of air that was once breathed by the great mathematician
Archimedes?
496. Median Mystery This is a bit of elementary geometry. A' and
B' are the mid-points of two sides, the lines AA' and BB' meet at X
and X divides each line in the ratio 2:1.
A
B ~ ~ - - - - - - - - ~ - - - - - - - - - - ~ C
156 Penguin Book of Curious and Interesting Puzzles
A
B ~ ~ - - - - - - - - ~ - - - - - - - - - - ~ C
A'
Suppose that B' is moved so that it divides CA in the ratio 2: 1, as
in the figure. How will the new intersection, Y, divide AA'?
497. Equal Products, Equal Sums It is obvious that 2 x 2 equals
2 + 2. However, 2 and 2 are also two equal numbers. What about
different numbers? Which set of different whole numbers has the
same product and the same sum?
498. The Island in the Lake The figure shows a small island, on
which is a tree, in the middle of a large and deep lake, which is 300
yards across. On the shore is another tree.
•
How might a man, who is unable to swim, with only a length of
rope rather more than 300 yards long, get from the shore to the
island?
499. A Present-able Poser Five pigeons are flying over a field in the
form of an equilateral triangle, of side 100 metres. Each pigeon, as it
The Puzzles 157
flies, makes a small deposit, as pigeons are wont to do. Explain why
at least one pair of these small deposits must be at most 50 metres
apart.
500. Five Points on a Lattice Five points are chosen on a square
lattice; in other words, five points of intersection are chosen on a
square grid. The figure illustrates just one possibtlity.
Why is it certain that at least one mid-point of a line joining a pair
of the chosen pOints, is also a lattice point?
501. Six on Five How can six matches be placed on a table so that
each of the matches touches all the other five matches?
502. The Last Match More and Less were playing a simple game.
They had a pde of twenty-one matches in front of them on the table
and they took turns to remove up to, but not more than, three
matches. The loser was the person who took the last match.
So far, Less has been less successful than More. Can you recom-
mend a strategy to Less which would make him more successful than
More, or at least guarantee that the games were spin more or less
evenly between them?
503. Squares and Triangles How can eight matches be placed so as
to form no less than two perfect squares and four tnangles?
504. How Many Triangles With three l i n e ~ only one mangle can be
created, With four Imes only fOUL How many can he created With SIX
straIght Imes?
158 Penguin Book of Curious and Interesting Puzzles
505. A Map-colouring Probiem It is easy to demonstrate that at
least four colours are needed to colour a map so that adjoining
countries are differently coloured.
This figure shows four countries, each of which borders the other
three. However, only three of the countries are the same shape. Can
you remedy this defect by drawing a map of four countries, of
identical shape and size, so that each borders the other three?
Langley's Adventitious Angles This tricky problem, named after E.
M. Langley, is famous because It is not as simple as it seems.
A

Tbe Puzzles 159
506. ABC is an isosceles triangle, whose vertex angle is 20°. DBC =
60° and ECB = 50°. All you have to do is to find angle BDE.
507. Heads over Tails Layout eight pen Illes in the circle, all heads
up. Start at any coin of your choice, count four as you touch four
coins in succession, and turn over a head. Choose one of the remaining
heads, count to four, starting with the chosen coin, and turn over the
fourth coin, to show tails up. Repeat until all but one of the coins are
tails up. Remember, you must start each time with a head, and the
fourth coin must be a head until you turn it tails up.
508. Imperfect Products 'Prove that the product of four consecutive
positive integers cannot be a perfect square.'
509. The Maximal Product 'What is the largest number which can
be obtained as the product of positive integers which add up to 100?'
510. The Programmer's Shirts 'A neat computer programmer wears
a clean shirt every day. If he drops off his laundry and picks up the
previous week's load every Monday night, how many shirts must he
own to keep him going?'
511. The Overlapping Squares Two squares are shown overleaf,
the larger being 10 inches square and the smaller being rather smaller.
A vertex of the second square lies at the centre of the first square, and
the centre of the second square lies on the right-hand edge of the first
square, one quarter of the way up the edge.
160 Penguin Book of Curious and Interesting Puzzles
What is the area of the overlap between the squares?
512. Cube Formation 'What is the shortest strip of paper 1 inch
wide and black on one side that can be folded to form a 1 inch cube
that is black on all sides?'
513. Delightful Discounts Buying from your favourite store you are
offered a discount of 5 per cent for payment in cash, 10 per cent as a
long-standing customer, and 20 per cent because it is sale time. In
what order should you take these discounts in order to pay as little as
possible for your purchase?
514. Fold and Fold Again Taking a large rectangular piece of thin
paper in your hands, you fold it in half once, and then in half again.
Repeating the same action, you fold it fifty times, each time in half.
After a few folds it is noticeably thicker. How thick is it after fifty
folds?
To be more precise, suppose that the original sheet is one-tenth of
a millimetre in thickness.
515. Pandigital Difference I thmk of a number which contains all the
digits 1 to 9, exactly once. I reverse it, so that the first digit becomes
the last, and find the difference between the numbers. The answer
also contains the digits 0 to 9. What number did I think of?
The Puzzles 161
516. Pandigital Square I think of a number which, curiously, re-
mains the same when I turn it upside down. When I square it, the
product contains all the 'digits from 0 to 9 exactly once each. What
number did I think of?
517. Folding a Square from a Rectangle 'You are given a rectangle
of paper ... the dimensions of which are unknown. You are required
to determine the side of a square which has an area equal to the
rectangle by merely folding the paper three times.'
518. The Square and the Triangle These five pieces can be as-
sembled to form a square and a triangle. How?
519. How Many Friends? At a party (meaning any gathering of
more than two people), at least two people will have the same
number of friends present - true or false?
520. Friends and Strangers At a small dinner party, which for the
purposes of this problem means a gathering of exactly six people,
there will always be either three people who are mutual friends, or
three guests who are mutual strangers. True or false?
521. Up and Down the Garden Path Lady Merchant's garden con-
sists of square plots of flowers surrounded by low box-hedges, with
paths between the plots.
Lady Merchant enters at the gate to the left and walks along the
paths to the summer house at the right-hand corner, every evening
taking, as far as possible, a different route.
162 Penguin Book of Curious and Interesting Puzzles
For how many successive days can she avoid repeating herself if
she is always moving towards the summer house?
522. Triangular Numbers This figure shows why the triangular
numbers were given their name by the ancient Greeks. Your puzzle is
to explain why every triangular number is the sum of a square
number and two other triangular numbers.
o o o o
o o o o o o
3
o o o o o o
6
o o o o
10
A further question: why is one more than every alternate triangular
number also the sum of a square and two triangular numbers?
The Puzzles 163
523. Tick and Cross There are twelve pentominoes, each composed
of five identical squares edge to complete edge. You will notice that
two of them resemble a V and an X, which we have called a Tick and
a Cross. How can an enlarged copy of each of these two figures, three
times as wide and three times as tall, be assembled each using nine of
the pentominoes?

z F L
p
N
524. Animals in the Cage Just as there are twelve distinct pentomi-
noes, so there are also, coincidentally, just twelve little 'animals' that
can be composed of six equilateral triangles fixed edge to complete
edge.
Crook

Signpost
Crown
IV\/\/
Bar
Snake

Butterfly
Hexagon 00
!¥7
Yacht
164 Penguin Book of Curious and Interesting Puzzles
Hook Lobster
Chevron ~ &. Sphinx
How can these twelve animals be packed into the rhombus which is
six units along each edge?
525. Locking Polyominoes Each of these polyominoes is composed
of twenty-five small squares, and you will notice that these two pieces
interlock, like pieces of a jigsaw puzzle, and also that they can be
used to tile the complete plane, continuing forever in every direction.
What is the smallest polyomino which will tile the plane, such that
each piece interlocks individually with each adjacent piece? And what
is the smallest polyomino tile if the condition is only that the
tessellation as a whole is interlocking, even if individual pieces are
not?
526. Zero Zeros How can 1,000,000,000 be written as the product
of two factors each of which contains no zeros at all?
The Puzzles 165
527. Two Children 'I have two children. They aren't both boys.
What is the probability that both children are girls?'
Now suppose that I have two children of whom the elder is a boy.
What is the probability that both are boys?
528. Pearls and Jars 'Mrs Tabako has fifty natural pearls, fifty
cultured pearls and two Ming jars. If she uses all the pearls, how
should she distribute them in the two jars in such a way that when
Mr Tabako enters the room and picks one pearl out of either jar at
random he will have the best possible chance of picking a cultured
pearl?'
529. The Chord in a Circle If a chord is drawn at random in a
circle, what is the probability that it will be longer than the side of
this equilateral triangle, inscribed in the circle?
530. Reptile Repeat It is quite easy - well, fairly easy - to cut this
rectangle-with-a-corner-misplaced into two identical pieces, as the
dotted lines show.
L ____ _
I
166 Penguin Book of Curious and Interesting Puzzles
Can you, however, cut it into three identical pieces? Three whole
pieces, that is; none of the pieces may be made up of smaller parts.
531. The Sphinx This shape, named for obvious reasons, can be cut
into four whole pieces, all identical in shape and all the same shape as
the original Sphinx. Curiously, the extra lines drawn total one half of
the perimeter of the original figure in length. How is it done?
532. Reproducing Reptile 'This is a reptile,' explained Peter, 'it can
be made up from identical smaller copies of itself. In fact,' he
continued, 'this shape has four sides and is made of four copies of
itself. '
'Really,' said Jane, 'that's very obvious,' and to show her disgust
she tore it into two halves, exclaiming, 'Each of these halves is a
much better reptile, because they are not symmetrical, and it is much
harder to see the answer.'
Sure enough, each half of the original shape was a reptile, divisible
into four copies of itself. What was the original shape?
533. Simple Sums Take any four-digit number, arrange the digits in
ascending and descending order to form two numbers, and subtract
the smaller from the larger. Repeat the same process with the answer.
What is the result - eventually?
534. Before the Invention of the Wheel A slab is being transported
on three rollers. If the circumference of each roller is 1 metre, how far
will the slab move as the rollers make one complete rotation?
535. The Most Ridiculous Route A postman with time to spare,
made a point of finishing his round by walking as far as possible
while visiting his last ten houses, which were equally spaced, 100
metres apart on a straight road. Starting at house No. 1 he delivered
The Puzzles 167
its mail and then walked to to, then back to 2, then all the way to 9,
and so on, zig-zagging up and down the road, and ending up at No.
6, where he was always offered a cup of tea and a bun, after walking
100 x (9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) = 4500 metres.
One day, however, it occurred to him that he might do 'worse'
than start at No. 1 and he planned an even longer route, which still,
however, ended up at No.6. What was it?
536. Bemusing Bolts Hold two identical bolts against each other, as
in the figure, and rotate them around each other, as if you were
'twiddling your thumbs'.
Will the bolts move apart or move closer together?
537. Packing Triangles I have two triangles, one larger than the
other. The longest, middle and shortest sides of the smaller are
shorter than the longest, middle and shortest sides of the larger, respec-
tively.
Can I be certain that the smaller triangle can actually be placed
inside the larger, without overlapping its edges?
538. Choosing in the Dark Miss Golightly is getting dressed in a
hurry, but the light in her closet has gone out. How many stockings
must she take from the stocking drawer, to ensure that she has a pair
of the same colour, if there are stockings of seven different colours in
the drawer?
539. Conway's Solitaire Army On an infinite square grid, an army
of men stand behind a starting line, waiting to move forward. Every
move consists of one man jumping over an adjacent man into the
empty square beyond, just as in solitaire. (The jumps may be made
horizontally, vertically or diagonally.)
168 Penguin Book of Curious and Interesting Puzzles
~
• • • • •
•
•
•
The figure, which is merely illustrative, shows how an army of
only eight men can send one man to the third rank beyond the
starting line.
Your problem, as General, is to decide just how far the army is
able to march. You are allowed, of course, to choose the size of the
army and to dispose your men in any manner you choose.
540. Points in a Square This is a square lattice. The pOints are all at
the vertices of identical squares, and you have to imagine, of course,
that the points are infinitely small. This is an essential point - pardon
the pun - because the problem is to decide whether it is possible to
draw a square on the lattice which contains exactly seventeen lattice
points in its interior and no lattice points on its perimeter.
• • • • • • •
• • • • • • •
• • • • • • •
• • • • • • •
• • • • • • •
• • • • • • •
• • • • •
The Puzzles 169
More generally, is there always a square which contains exactly N
lattice points in its interior, where N is any integer you choose?
541. The Squirrel and the Hunter A hunter sees a squirrel in a tree,
and walks towards it. As he does so, the squirrel disappears round
the far side of the trunk, and as the hunter circles the tree the squirrel
keeps out of sight on the other side, circling also. As this curious
chase continues there is no doubt that they are both circling the tree,
but, are they circling each other?
William James, the famous psychologist, posed this problem in his
book Pragmatism. What is your pragmatic response?
542. Farthing Fiddle One of the advantages of the decimal system
of coinage is that any sum of pence can instantly be converted to
pounds by inserting a decimal point. In the days of pounds, shillings
and pence this was not possible - in general. However, there was one
five-figure quantity of farthings which could be converted into £.s.d.
by simply inserting two strokes of the pen.
What was it? You may recall that £1 = 20 shillings = 240 pence,
and a farthing is a quarter of one penny.
543. To Knot or not to be Knotted? Glancing at this loop of string
on a table, too quickly to notice which bits go over which other bits,
you idly ask yourself whether it is likely to be knotted? What is the
answer?
170 Penguin Book of Curious and Interesting Puzzles
544. Cooked Turkey 'An old invoice showed that seventy-two tur-
keys had been purchased for "-67.9-". The first and last digits were il-
legible.'
How much did each turkey cost?
545. The Chauffeur Problem 'Mr Smith, a commuter, is picked up
each day at the train station at exactly 5 o'clock. One day he arrived
unannounced on the 4 o'clock train and began to walk home.
Eventually he met the chauffeur driving to the station to get him. The
chauffeur drove him the rest of the way home, getting him there 20
minutes earlier than usual.
'On another day, Mr Smith arrived unexpectedly on the 4.30 train,
and again began walking home. Again he met the chauffeur and rode
the rest of the way with him. How much ahead of usual were they
this time?'
546. An Express Problem An express train takes 3 seconds to enter
a tunnel which is 1 km long. If it is travelling at 120 km an hour, how
long will it take to pass completely through the tunnel?
547. The Lost Paddle 'A man went upstream from his dock In a
motorboat. As he passed under a bridge one mile from the dock his
emergency paddle fell overboard, a loss which he did not discover
until 10 minutes later, whereupon he went back downstream to
retrieve his paddle, and caught up to it directly opposite his dock. If
he travelled at constant water speed and lost no appreciable time
turning round, what was the rate of the current of the river?'
548. Spot the Blunder Puzzle-solvers must be wide awake to solu-
tions which seem to be solid as a rock but actually contain large
holes. Here is a puzzle and the solution as published in a book which
shall be nameless. Spot the boob!
A spy is watching the Pentagon, which as you know is a large
building in the form of a regular pentagon, from a distance with
powerful binoculars. What is the chance that he or she can see three
sides of it?
This is the offered solution: Imagine another spy at an equal distance
away, exactly opposite to the first spy. If one spy can see two Sides,
the other will see three. Since It is equally likely that the spy will be at
either spot, the probability is one-half.
The Puzzles 171
549. The Burning Candles 'On Christmas Eve two candles, one of
which was one inch longer than the other, were lighted. The longer
one was lighted at 4.30 and the shorter one at 6.00. At 8.30 they were
both the same length. The longer one burned out at 10.30, and the
shorter one at to.OO. How long was each candle originally?'
550. The Heavy Boxes 'Five equal cubical boxes, each with an A on
its top side, stand together as in the first figure.
A
A A A
A
'The boxes are to be brought in.o line, but they are so heavy that
they can be moved only by tipping them over about an edge. With
these conditions, it proves to be impossible to bring them into line
with all the A's the same way up, and the arrangement finally
achieved has the plan view shown in the second figure. Which box
was orIginally in the middle?'
A A A A
551. Passing Trains A man standing on a platform notes that a
train going in one direction takes 3 seconds to pass him, and a train of
the same length in the other direction takes 4 seconds. How long did
it take for them to pass each other?
172 Penguin Book of Curious and Interesting Puzzles
552. Triangles in a Triangle How many triangles can be counted in
this figure?
553. The Prisoner's Dilemma Because he is deemed to be a foolish
man who has allowed himself to be led into crime by his companions,
the prisoner has been given a last chance. He is shown two doors in
the courtyard, one of which leads to freedom and the other to a long
sentence. Each is guarded by a warder, one of whom always lies and
one of whom is impeccably honest, but he does not know which is
which.
He is allowed one question, to be put to one of the warders. How
can he discover which is the door to freedom?
554. Short-list
'1. The number of the first true statement here added to the number
of the second false statement gives the number of a statement
which is true.
2. There are more true statements than false.
3. The number of the second true statement added to the number of
the first false statement gives the number of a statement which is
true.
4. There are no two consecutive true statements.
5. There are at most three false statements.
6. If this puzzle consisted of statements 1 to 5 only, then the answer
to the following question would still be the same.
Which statements are true?'
The Puzzles 173
555. The Crossed Cylinders Two identical cylinders are placed so
that their axes cross at right-angles and their common volume, has
four identical curved surfaces.
How can the volume of this common solid be calculated without
the use of any calculus?
556. Concyclic Points 'Five paper rectangles - one with a corner
torn off - and seven paper disks have been tossed on a table. They lie
as shown in the figure. Each corner of a rectangle and each spot
where edges intersect makes a point. The problem is to find three sets
of four concyclic points: four points that can be shown to lie on a
circle. For example, the corners of rectangle R are such a set, because
the corners of any rectangle lie in a circle. What are the other two
sets?'
174 Penguin Book of Curious and Interesting Puzzles
557. Hot Cross Buns 'The hot cross bun man cried:
Hot cross buns, hot cross buns,
One a penny, two a penny, hot cross buns,
If your daughters don't like them
Give them to your sons!
Two a penny, three a penny, hot cross buns,
I had as many daughters as I have sons
So I gave them seven pennies
To buy their hot cross buns.
How many children were there if they were all treated alike and if
there was only one way in which to purchase the buns?'
558. White to Play This looks like the position after White has
made a rather unusual first move, yet it is, in fact, the position after
Black has just played.
What is the smallest number of moves that could have been played, in
order to reach this position with White to play?
The Puzzles 175
559. The Unwound Clock 'I have no watch, but I have an excellent
clock, which I occasIOnally forget to wind. Once when this happened
I went to the house of a friend, passed the evening in listening to a
radio concert programme, and went back and set my clock. How
could I do this without knowing beforehand the length of the trip?'
560. Tom's House 'John is trying to find out where Tom lives, and
all he knows is that it is in a street where the houses are numbered
from 8 to 100 (inclusive). John asks, "Is it greater than 50?" and Tom
answers, but lies. John then asks, "Is the number a multiple of 4?"
Again Tom answers, and again he lies. Then John says, "Is it a
perfect square?" Tom answers and this time he tells the truth. Finally
John asks "Is the first digit 3?" After Tom has replied (truthfully or
not we do not know!) John tells him the number. He is wrong! What
was the number of Tom's house?'
561. The Same Sister Is it possible for two men who are completely
unrelated to each other, to have the same sister?
562. C is Silent 'On the Island of Imperfection there are three
tribes, the Pukkas, who always tell the truth, the Wotta-Woppas,
who never tell the truth, and the Shilli-Shallas, who make statements
which are alternately true and false, or false and true.
'As the reader can imagine, it is always the most important part of
life on the island to discover to which tribe people belong. On a
recent visit I was doing some work on this with three inhabitants
whom I shall call A, Band C. They have got into the habit lately of
going around in threes, one from each tribe, and I am glad to say that
these three were no exception.
'C did not make my self-appointed job as a detective any easier by
being silent, but the other two spoke as follows:
A: "C is a Pukka."
B: "A is a Pukka."
'Find the tribes to which A, Band C belong.'
563. A cylinder can be 'squared' with the usc of only ten square pieces.
How can squares of edges 30, 27, 25, 17, 15, 13, 11, 8, 3 and 2 be
fitted together to fill the space between two parallel lines, in such a
way that when the opposite edges are joined, it forms a 'squared'
cylinder?
176 Penguin Book of Curious and Interesting P.uzzles
564. Speaking of Bets ' "The three of us made some bets.
1. First, A won from B as much as A had originally.
2. Next, B won from C as much as B then had left.
3. Finally, C won from A as much as C then had left.
4. We ended up having equal amounts of money.
5. I began with 50 cents."
'Which of the three - A, B or C - is the speaker?'
565. The Murderess 'Three women, named Anna, Babs and Cora,
were questioned about the murder of Dana. One of the three women
committed the murder, the second was an accomplice in the murder,
and the third was innocent of any involvement in the murder.
'Each of the following three statements was made by one of the
three women:
1. Anna is not the accomplice.
2. Babs is not the murderess.
3. Cora is not the innocent one.
I. Each statement refers to a woman other than the speaker.
II. The innocent woman made at least one of these statements.
III. Only the innocent woman told the truth.
Which one of the three women was the murderess?'
566. Six Gs In the multiplication problem below, each letter repre-
sents a different digit:
ABCDE
x F
GGGGGG
Which of the ten digits does G represent?
567. The Wheels of Commerce '''How's the motor business?" asked
Bob, glancing at the menu.
'Ben owns a used car lot. His cars are good; his prices are right; his
guarantee means just what it says. Other dealers come and go, but
Ben keeps right on selling. "Not too bright around Christmas," he
replied, "but sales have picked up again."
'''That's dandy!" commented Bob. "I was talking to Stan Logan
The Puzzles 177
down on Wardie and Myrtle yesterday. He's hardly sold a car this
year."
'Ben smiled. "A lot of them are having a tough time," he said, "but
maybe I'm lucky. We've done well so far this month - each week
more sales than the previous week."
'''What's that in actual numbers?" asked Bob, who's a great one
for facts.
'''I'm not sure about the last few days," replied Ben, "but we sold
fifty-six cars the first three weeks. And here's something to amuse
yourself with." He thought a moment. "The difference between the
numbers we sold in the first and second weeks, multiplied by the
difference between the second and third weeks, comes to the same as
the number we sold the first week."
'The shapely waitress leaned over his friend just then to take their
order, and Bob rather lost interest in car sales. But how many cars
would you say Ben sold in the third week?'
568. Professor Mesozoic, the famous geologist, had a problem. She
had lost her notes on the samples of sedimentary rock that she had
collected, and she no longer knew in what orientation they had been
found. Then her assistant, Slatebed, had a bright idea. Within the
rock were numerous tiny specks of a mineral, which it was reasonable
to suppose had been randomly distributed in the material when it was
laid down on the bed of some ancient lake. Subsequently, as the
sediment was vertically compressed into rock, they would have been
forced together in that vertical direction, and examination of their
present distribution would show what that direction had been!
He explained his idea enthusiastically, but Professor Mesozoic
thought for a moment, and then rejected his idea. Why?
The Solutions
1. 7 + 49 + 343 + 2401 + 16,807 = 19,607.
2. 7 + 49 + 343 + 2401 + 16,807 + 117,649 = 137,256.
[Boyer, 1985, p. 210]
3. One! All the others were coming from St Ives!
[Midonick, 1965, quoting Every Child's Mother Goose, introduc-
tion by Carolyn Wells, New York, 1918]
4. Each of the five solutions has nine terms. 1 = 113 + 115 + 117 +
119 + 1111 + 1115 + 1135 + 1145 + 11231 has the smallest larger
denominator, 231.
[Gardner, 1978a]
5. The smallest value of the denominator is 25: the greedy algorithm
gives 3/25 = 119 + 11113 + 1125425, but 3/25 also equals 1110 +
1150.
[Gardner, 1978a]
6. 112 = 1/22 + 113
2
+ 1142 + 1/5
1
+ 1/7
1
+ 11121 + 1/15' +
1120
1
+ 1128
1
+ 1135
2
.
[Szurek, 1987, p. 391]
7. 9.
[Peet, 1923, p. 63]
8. 13z\.
[Peet, 1923, p. 65]
9. 16 + 1/56 + 11679 + 11776. This and similar problems were
solved by the rule of false position. An answer which was judged to
182 Penguin Book of Curious and Interesting Puzzles
be roughly correct was chosen, and then adjusted by multiplying by a
suitable factor.
In this case the scribe guessed 16 immediately, and got the wrong
but close answer 36 + 2/3 + 114 + 1128, which falls short of 37 by
2/42. Next the scribe calculates that 1 + 2/3 + 112 + 117 is 97/42 so
that the multiplier required is 2/97, which the scribe could read off
from the earlier table in the Rhind papyrus of fractions 2/n, without
any further calculation. It is 1156 + 11679 + 11776.
[Peet, 1923, p. 69]
10. 11 + lOB + 20 + 291 + 38i = 100. This problem was also solved
by false position. The scribe first artificially constructs the series 1 +
6t + 12 + 17t + 23 = 60, which has the property that the first two
terms sum to one-seventh of the last three. Each term is then multi-
plied by Ii to change the 60 into 100.
[Peet, 1923, p. 78]
11. 6
2
+ 8
2
= 100.
[Gillings, 1972, p. 161]
12. The difference between each successive share is 37,55 or 37t\.
[Neugebauer and Sachs, 1945, p. 53]
13. tOOO = 10' + 30'.
[Eves, 1976, p. 46]
14. The distance is the third side of a right-angled triangle with
hypotenuse 0,30 and one leg 0,30 - 0,6 = 0,24. The third leg is
length 0,18.
[Eves, 1976, p. 46]
15. Given that a, band c are integers, such that a' + b' = c', either
a or b is even; suppose that a is even.
Then there are integers p and q such that a = 2pq, b = p' - q'
and c = p2 + qZ.
The evidence that the Babylonians used this formula is simple: the
values of p and q which fit the numbers on Plimpton 322 are all so-
called 'regular' numbers whose factors are powers of only 2, 3, and 5
or products of such powers.
The ratio cia = t(plq + qlp). The Babylonians could now find
suitable values of p and q by referring to the standard reciprocal
tables which they used for multiplication anyway. We lack such
The Solutions 183
tables, so to make this equal to approximately 1.54, put p/q = t and
solve the quadratic!(t + lit) = 1.54:
t
2
+ 1 = 3.08t
t ~ 2.711 or 0.369
We discard 0.369, as we want p > q:
Take
P 27
- = - as a rough approximation, so that
q 10
Then
and
P = 27, q = 10
a = 2pq = 540
c = pI + qI = 829
c
1.535
a
The approximate ratio 1.54 in the problem was taken from the
figures 3541 and 2291 in Plimpton.
[Neugebauer and Sachs, 1945, pp. 38-41; Eves, 1976, p. 37)
16. Let the letters X, Y, Z and T denote the numbers of white, black,
dappled and yellow bulls respectively, and x, y, z and t denote the
number of white, black, dappled and yellow cows, respectively. Then
the conditions of the problem give seven equations in these eight
unknowns:
(1) X - T = 5/6 Y
(2) Y - T = 9/20 Z
(3) Z - T = 13/42 X
(4) x = 7/12 (Y + y)
(5) y = 9/20 (Z + z)
(6) z = 11/30 (T + t)
(7) t = 13/42 (X + x)
From the first three equations, X, Y and Z can be found in terms of
T:
X = 742/297 T Y = 178/99 T Z = 1580/891 T
Since 891 and 1580 possess no common factors, T must be some
whole multiple -let us say G - of 891. Consequently,
X = 2226G Y = 1602G Z = 1580G T = 891G
If these values are substituted into equations (4), (5), (6) and (7), the
following equations are obtained:
12x - 7y = 11214G 20y - 9z = 14220G
184 Penguin Book of Curious and Interesting Puzzles
30z - llt = 9801G 42t - 13x = 28938G
These equations are solved for the four unknowns x, y, z and t and
we obtain:
4657 x = n06360G
4657z = 3515820G
4657y = 4893246G
4657t = 5439213G
m which the number 4657 is prime. Since it divides none of the
coefficients on the right, 4657 must divide G. Taking the simplest
case, let G = 4657, we obtain as the smallest solution:
white bulls
black bulls
dappled bulls
yellow bulls
10,366,482
7,640,514
7,358,060
4,149,387
white cows
black cows
dappled cows
yellow cows
7,206,360
4,893,246
3,515,820
5,439,213
[Archimedes' cattle problem, as taken from T. L. Heath, The
Works of Archimedes with the Method of Archimedes, Dover,
n.d., p. 319. This solution follows Dome, 1965]
17.
18. Suppose that Mary alms for point P on the river bank. Reflect
Mary's original position in the line of the river bank. Then the
distance SPT equals the distance S'PT and the latter will be a
mimmum when S'PT is a straight line. It follows that P IS the point
such that SP and PT make the same angle with the line of the river.
Heron used exactly the same argument by reflection to conclude
that when light is reflected, the angles of incidence and reflection are
equal. This is one of the earliest solutions to an extremal problem. As
The Solutions 185
s·
one Greek commentator remarked on Heron's solution, expressing a
view which has haunted and inspired scientists ever since, 'for Nature
does nothing in vain nor labours in vain.'
19. 'My right eye fills 118 jar in 6 hours [taking a day to be 24 hours,
where the Greeks might have taken it to be 12], and my left eye fills
1112 in 6 hours, and my foot 1116. Thus all four fill the jar 1 +
118 + 1112 + 1116 = Iii times in 6 hours. So the jar will be filled once
in 6 x 48/61 hours, or 47 minutes and 13 seconds, approximately.'
[Sandford, 1930, p. 216]
20. A, Band C do the whole work in 10, 15 and 30 days, respec-
tively.
21. Suppose that the given ratio is n rather than 3. Then, if u,v and
x,y are the sides of two such rectangles, the equations can be written:
u + v = n(x + y) xy = nuv
and Heron's solution runs parallel to the general solution:
x=2n
3
-1
u = n(4n3 - 2)
y = 2n
3
v=n
which leads Heron to his solution: the rectangles are 53 x 54 and
318 x 3.
[Thomas, 1980, p. 505]
186 Penguin Book of Curious and Interesting Puzzles
22. There is no very simple solution to this problem. The sides are
20,21 and 29, and the area is 210.
[Thomas, 1980, pp. 507-8)
23. 20.
24. The first number is 98, the second, 94.
25.9,7,4andl1.
[The quotation from Xylander is from Ore, 1948, p. 195)
26. 6, 4 and 5 is the simplest solution, but it is the ratios of the
numbers which are important, so, for example, 12, 8 and 10 is
another solution.
27. Diophantos's answer is 1,7 and 9.
This is his solution, which illustrates very well his methods, which
tend to simplify the problem and produce one or a few solutions. It
was left to later Indian mathematicians to find more general solu-
tions.
'Take a square and subtract part of it for the third number; let
Xl + 6x + 9 be one of the sums, and 9 the third number.
'Therefore product of the first and second = Xl + 6x; let first =
x, and that second = X + 6.
'By the two remaining conditions, lOx + 54 and lOx + 6 are both
squares.
'Therefore we have to find two squares differing by 48; this is easy
and can be done in an infinite number of ways.
'The squares 16, 64 satisfy the condition. Equating these squares to
the respective expressions, we obtain x = 1 and the numbers are 1, 7,
9.'
28. This is Diophantos's solution:
'Let the sum of all three be Xl + 2x + 1, sum of first and second
Xl, and therefore the third 2x + 1; let sum of second and third be
(x - 1)'.
'Therefore the first = 4x, and the second = Xl - 4x.
'But first + third = square, that is, 6x + 1 = square = 121, say.
'Therefore x = 20 and the numbers are 80, 320, 41.'
The Solutions 187
29. 'Let x = the whole number of measures; therefore Xl - 60 was
the price paid, which is a square = (x - m)l, say.
Now t of the price of the five-drachma measures + 1 of the price of
the eight-drachma measures = x;
so that Xl - 60, the total price, has to be divided into two parts
such that t of one + 1 of the other = x.
We cannot have a real solution of this unless x > 1(x
1
- 60) and
<t(x
1
- 60).
Therefore 5x < Xl - 60 < 8x.
(1) Since x' > 5x + 60,
Xl = 5x + a number greater than 60,
whence x is not less than 11.
(2) x' < 8x + 60
or Xl = 8x + some number less than 60,
whence x is not greater than 12.
Therefore 11 < x < 12.
Now (from above) x = (m' + 60)/2m;
therefore 22m < m
1
+ 60 < 24m.
Thus (1) 22m = m
1
+ (some number less than 60),
and therefore m is not less than 19.
(2) 24m = m' + (some number greater than 60),
and therefore m is less than 21.
Hence we put m = 20, and
Xl - 60 = (x - 20)',
so that x = lIt, x' = 1321, and x' - 60 = 721.
Thus we have to divide 721 into two parts such that t of one part
plus 1 of the other = 1 H.
Let the first part be 5z.
Therefore i (second part) = It! - z,
or second part = 92 - 8z;
therefore 5z + 92 - 8z = 721;
and z = 79/12.
Therefore the number of five-drachma measures = 79/12.
Therefore the number of eight-drachma measures = 59/12.'
[Midonick, 1968, pp. 48-9]
30. The string must form a semi-circle. Imagine that it takes the form
in the figure, and reflect the shape in the shore-line. Then the entire
closed curve will be the curve that encloses the largest area, for
double the length of string. This is a circle, a fact which follows from
the theorem that the polygon with a given number of sides, with
188 Penguin Book of Curious and Interesting Puzzles
\ I
\ I
\ I
, I
" I
',......... ///
............. _-------" ....
maximum area, is a regular polygon, if the number of sides is then
allowed to tend to infinity.
31. The area is a maximum when the ends of the rods lie on a circle.
This conclusion is suggested by the thought that if the quadrilateral is
adjusted so that its vertices do lie on a circle, which is certainly
possible, and the four arcs of the circle are then also hinged at the
vertices of the quadrilateral, and the figure moved, then the area
surrounded by the four circular arcs cannot be a maximum since they
no longer form a circle: yet the areas between the arcs and the sides
of the quadrilateral have not changed - only the interior area of the
quadrilateral changes when the figure is moved about its hinges.
The area can be calculated by a formula that was discovered by
Brahmagupta but also apparently known to Archimedes. If half the
sum of the sides, a, b, c and d, is s, then the area is given by
A = J(s - a)(s - b)(s - c)(s - d)
(If one of the sides has zero length, then the quadrilateral becomes a
triangle, which is automatically inscribable in a circle, and this
formula becomes Heron's formula ~ ) ( s - b)(s - c) for the
area of a triangle with sides a, band c; s is half the of the sides.)
32. Reflect the original figure in both walls, and then reflect a third
time, to get this complete figure.
The area enclosed by the screen will be a maximum when the area
of the entire octagon is a maximum, and this will be so when it is a
regular octagon. So the screen must be placed so that it meets the
walls at two angles of 67}0 each.
The Solutions 189
33. Reflect the isosceles triangle in its third, variable, side, to form a
rhombus. The area of the rhombus will be a maximum when it is a
square, and so the area of the isosceles triangle is a maximum when
the angle between its equal sides is a right-angle.
34. 5 ~ hours are past and ~ remain.
35. He was a boy for fourteen years, a youth for seven; at 33 he
married, and at 38 he had a son born to him who died at the age of
42. The father survived him for four years, dying at the age of 84.
36. 577; and 422;.
[Problems 34-6 are from The Greek Anthology, 1941]
37. If the estate remaining after payment of the legacy is divided into
twenty parts, the husband receives five, the son six and each daughter
three. The stranger receives 15/56, so Al-Khwarizmi divides the
whole estate into 20 x 56 = 1120 parts. The stranger receives 300,
the husband 205, the son 246, and each daughter 123.
38. Abul Wafa gave five different solutions. Here are three of them.
Let one vertex of the equilateral triangle be at D. Construct N so
that ABN is equilateral. Mark F on AB so that AB = BF, and draw
an arc cutting ABF so that FN = FG. Then G is one of the other
vertices, and the last vertex is easily found on BC
190 Penguin Book of Curious and Interesting Puzzles
F
Construct AID to be equilateral. Bisect AD! and then bisect the
half towards AD again. The second bisector cuts AB at G, one of the
other vertices sought.

Join B to the mid-point, M, of DC. Draw an arc with centre Band
radius BA to cut BM at N. Let DN cut CB at H. Then H is one of the
other vertices sought.

[Berggren, 1986]
The Solutions 191
39. Abul Wafa's solution bisects two of the squares and places them
symmetrically around the third. Joining vertices by the dotted lines,
the larger square is found. The four small pieces outside it fit exactly
the spaces inside its boundary.
This solution is not as idiosyncratic as it initially appears. If one of
the original squares plus a square composed of all four of the
'quarters' are repeated to form a tessellation, then joining the dotted
lines is one standard way of one small square and one large
square into a single square. It also works if the larger square is
thought of as composed of four quarters.
Note that the simplest dissection of a Greek Cross into a square
(see p. 228 below) can also be seen as an example of Abul Wafa's
theme.
40. The same solution works if the two larger squares are bisected.
Their size is irrelevant.
192 Penguin Book of Curious and Interesting Puzzles
41. Dissect the larger hexagon into six identical triangles, as here,
and then arrange them around the smaller hexagon. The dotted lines
complete the dissection.
[Wells, 1975]
42. Arrange the three larger triangles round the small similar tri-
angles, like this, and join the vertices as indicated.
[Wells, 1975]
43. Mark AC equal to the fixed radius, and draw two arcs, with
centres A and C, to construct D, the third vertex of equilateral
triangle ACD. Extend the line CD and mark off E, so that DE = CD.
Then AE is perpendicular to AB.
The Solutions 193
E
A c B
44. Construct perpendiculars in opposite directions at the endpoints
A and B. Mark off as many segments as necessary along each
perpendicular, using the fixed radius. Joining the first mark on one
perpendicular to the nth mark on the other, and so on, will divide the
segment AB into n + 1 equal parts.
194 Penguin Book of Curious and Interesting Puzzles
45. 'At the endpoint of A of the radius DA, erect AE perpendicular to
AD and on AE mark off AE = AD, then bisect AD at Z and draw
the line ZE.
'On this line mark off ZH = AD and bisect ZH at T. Then
construct a line through T, perpendicular to EZ, and let it meet DA
extended at I.
'Finally, let the circle with centre I and radius AD meet the given
circle at points M and L.'
4 - 4 - - - ~ - - ~ ~ - - - - - - ~ O
M and L are two vertices of the required pentagon, 0 is another,
and the perpendicular bisectors of MO and LO meet the circle at the
other vertices.
[Berggren, 1986]-
46. Sissa required 2'· - 1 = 18,446,744,073,709,551,615 grains of
wheat. Cassell's Book of Indoor Amusements, Card Games and
Fireside Fun (1881) calculates, taking an average number of 9216
wheat kernels to a pint, that this is a total of 31,274,997,411,298
bushels of grain, 'a larger amount than the whole world would
produce in several years'.
By way of a modern example, the authors further calculate that if
one pin were dropped into the hold of the Great Eastern steamship
(22,500 tons) in the first week, two in the next and so on, a year's
worth would fill 27,924 Great Easterns.
As Kasner and Newman remark, this is the same as the number of
moves required to transfer all the rings in Lucas's 'Tower of Hanoi'
puzzle (problem 238) and also roughly the number of ancestors that
The Solutions 195
each person alive today had at the start of the Christian era, which
happens to be about sixty-four generations ago. The ratio of 2'4 to
the actual population of the earth at that time is therefore a measure
of the amount of unintentional interbreeding that has taken place.
[Kasner and Newman, 1949]
47. Let the numbers of men, women and children be m, wand c
respectively. Then the problem states that
m + w + c = 20 and 3m + !w + ic = 20
It follows that 5m + 2w = 20 and the unique solution IS m = 2,
w = 5, and c = 13.
48. Fifty soldiers broke down and fifteen remained in the field.
[Mahavira, 1912, p. 112]
49. If bla and dlc are the original selling prices, then the average
price is Hbla + die). The trusts set the price to be (b + d)/(a + c).
Comparing these two expressions, and simplifying, it follows that the
trust price will be advantageous only if a > c and bla > dlc, that is,
if the original prices are unequal and the denominator of the higher
price is greater than that of the lower price.
[The theme of this problem occurs in Mahavlra; this version is
taken from Kraitchik, 1955, pp. 35-6]
50. After: and! have reached the maid-servant and the bed, one half
remain. These are halved again and again, six times in all, leaving
11128 = 1161. The total number of pearls is therefore the improbable
148,608.
[Mahavira, 1912, p. 73]
51. There are six ways of choosing a single flavour, and (6 x 5)/2 =
15 ways of choosing a pair of flavours. Similarly there are (6 x 5 x
4)/(3 x 2) = 20 ways of choosing three flavours, and (6 x 5 x 4 x
3)/(4 x 3 x 2) = 15 choices of four flavours. This last figure is equal
to that for a choice of two flavours because choosing four flavours is
the same as choosing two whIch you will not include. By the same
reasoning there are six ways of choosmg five flavours.
The total of all these answers, including the single way in which all
the flavours can be rejected is 2 x 2 x 2 x 2 x 2 x 2 = 64, because
each flavour can be either rejected or accepted.
[Mahavira, 1912, p. 150]
196 Penguin Book of Curious and Interesting Puzzles
52. Let the value of the purse be x, and the wealth of the three
merchants, p, q and r. From the equations
p + x = 2q + 2r
q + x = 3p + 3r
r + x = 5p + 5q
it follows that p: q: r = 1: 3: 5, and the solution in smallest integers is
that the merchants originally had 1,3 and 5 in money, and the value
of the purse was 15.
53. The number of arrows in a bundle is the sum, as far as necessary,
of the series 1 + 6 + 12 + 18 + ...
If eighteen arrows are visible, there are thirty-seven arrows in all.
[Mahavira, 1912, p. 167]
54. Mahavira does not state the distance between the pillars because
this need not be known. The height reqUired is one half of the
harmonic mean of the given heights, that is, if the heights of the
pillars are P and Q, then the required height is
or
1
PQ
(P + Q)
(The point at which the string touches the ground divides the horizon-
tal distance between the pillars in the ratio of their heights.)
The Solutions 197
Mahavira also solves the problem in which the strings are attached
to the ground at points outside the bases of the pillars.
[Mahavira, 1912, p. 243]
55. The common difference is 22/7.
[Eves, 1976, p. 199; Midonick, 1965, p. 277]
56. From the figure, X, = 15
'
+ (45 - Xl' from which x = 25. Alert
readers might spot at once that the triangle is just the enlarged 3--4-5
triangle.
[Eves, 1976, p. 199]
57. 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3,628,800 pos-
sible arrangements of Siva's attributes.
There are 4 x 3 x 2 x 1 = 24 arrangements of Vishnu's, all of
which, according to Midonick, have their own special names.
[Sandford, 1930, p. 198; Midonick, 1965, p. 275]
58. Let the value of each blue gem be b, of each emerald, e, and of
each diamond, d. Then,
12b + 2e + 2d = 6e + 2b + 2d = 4d + 2e + 2b
It follows that the ratios b: e: dare 2: 5 : 10, and these are the
simplest possible integral values for their worth, which cannot be
determined more exactly.
198 Penguin Book of Curious and Interesting Puzzles
59. This is the original Lo Shu diagram. The solution is essentially
unique, but there are eight possible solutions obtained by merely
reflecting and rotating anyone of them.
4
~ O O O D o a o o o /
I + I
3
o 8
[Needham, 1959, p. 57]
60. Twenty-three pheasants and twelve rabbits.
[Midonick, 1965, p. 183]
61. Seven men and fifty-three articles.
9
5
1
[Midonick, 1965, p. 183; Mikami, 1964, p. 16]
62. Each ox, 1* tael; each sheep, ¥f tae!'
[Midonick, 1965, p. 183]
63. 91,41,21 measures of grain, respectively.
[Mikami, 1964, p. 18]
2
7
6
64. By Pythagoras, or rather by the Gougu theorem, which was the
Chinese name for the theorem of Pythagoras, the water is 12 feet
deep. The right-angled triangle formed by the length of the reed
excluding the protruding foot, the line from its base to the edge of the
pool, and the line from that same point on the edge to the point
where the reed breaks the water, is a 12-13-5 triangle.
However, the only requirement of the problem is that the hypot-
enuse be 1 unit longer than another leg. In general, the formula (2n +
1)1 + (2n
l
+ 2n)' = (2n
'
+ 2n + 1)' gives a right-angled triangle
with hypotenuse and one leg differing by unity.
[Midonick, 1965, p. 184]
The Solutions 199
65. 17 feet, by Pythagoras: the figure forms an 8-15-17 triangle.
[Midonick, 1965, p. 184]
66. 4¥0 feet. If the height of the break is x, then x' + Y = (10 - x)'.
10
x
3
[Midonick, 1965, p. 184]
67. The circle has diameter 6. Liu Hui, the author of the Sea-Island
Arithmetical Classic, demonstrated this solution by a dissection.
A
The triangle containing the inscribed circle is doubled to form a
rectangle, of dimensions a x b where a, band c are the shorter sides
and hypotenuse of the original triangle, respectively.
200 Penguin Book of Curious and Interesting Puzzles
The pieces are then rearranged to form a rectangle of height D, the
diameter of the required circle, and length a + b + c. It follows that
the required diameter is given by D = 2ab/(a + b + c). Here a = 8,
b = 15 and c = 17.
[Li Yan and Du Shiran, 1987, p. 71]
68. 2f., days, when both grow to the same height of 4ll feet.
[Mikami, 1964, p. 18]
69. There were sixty guests. The rule given by Sun Tsu is 'Arrange
the 65 dishes, and multiply by 12, when we get 780. Divide it by 13,
and thus we obtain the answer.' This follows from the fact that if
there were x guests, then I + J + i = 65.
[Mikami, 1964, pp. 31-2]
70. The numbers in this sequence all leave a remainder 2 when
divided by 3: 2,5,8, 11, 14, 17,20,23,26, ... Of these numbers, the
sequence of numbers 8, 23, 38, ... , also leave a remainder of 3 when
divided by 5. Out of this sequence, the numbers 23, 128, 233, ...
(where the difference is always 3 x 5 x 7 = 105) also leave a remain-
der of 2 when divided by 7.
Therefore the smallest possible number of unknown things is 23,
but there are in fact an infinite number of solutions to the original
problem, i.e. all the numbers in the final sequence.
71. Sixty days, the lowest common multiple of 3,4 and 5.
[Mikami, 1964, p. 33]
72. If the lengths of the shorter sides are a and b, the side of the
square is ab/(a + b), as Liu Hui proved by this beautiful figure, similar
to the figure he used to solve problem 67:
The Solutions 201
A
---T-----..,
1 1
1 1
I 1
1 1
1 1
- ____ -1
I
1
~ - - - - ~ - - - ~ - - - - - - - - - ~ - - - - - ,
1 ","1 1
I ".'" I I
I ," I I
I.,' I I
~ . ~ ____ ~ __ ~ ~ I ~ ________ L _____ J
cl_ ~ I - ~ I
The original diagram is repeated to complete a rectangle a x b,
and this is then reassembled to form an equal rectangle of length a +
b and height equal to the side of the required square.
[Li Yan and Du Shiran, 1987, p. 70]
73. He catches lip 37 - 23 = 14 miles in 145 miles, so he will catch
up 37 miles in 145 x 37/14 = 383t.. miles.
[Mikami, 1964, p. 41]
74. If the numbers of cocks, hens and chicks are c, hand 5 respec-
tively, then the conditions are:
c + h + 5 = 100
5c + 3h + 15 = 100
which together imply 7c + 4h = 100.
These equations are indeterminate - there are not enough condi-
tions to fix the values exactly. However, given that only whole
numbers of birds were sold, and some were sold of each kind, there
are just three solutions:
c = 12
c = 8
c=4
h=4
h = 11
h = 18
5 = 84
5 = 81
5 = 78
If the number of kinds of articles sold is increased, but with the
same information given, then the number of possible solutions also
increases dramatically. The Arabic author Abu Kamil, in his Book of
Arithmetical Rarities, written just before 900 AD, supposed that 100
birds are sold for 100 drachma, the birds being ducks at 2 drachma,
202 Penguin Book of Curious and Interesting Puzzles
hens at 1 drachma, doves 2 for a drachma, ring-doves 3 for 1
drachma and larks 4 for 1 drachma. He wrote, 'I went into [this
problem] fully and found that there were 2,696 valid answers. 1
marvelled at this, only to discover - when 1 spoke of it - that 1 was
reckoned a simpleton or an incompetent, and strangers looked upon
me with suspicion. So 1 decided to write a book ... '
This must be one of the better reasons for writing a book. The
actual number of solutions is 2,678, and Abu Kamil is probably the
first mathematician to have considered the number of solutions of a
problem as a feature in itself. At that time, it was usually considered
adequate to find one solution, following the example of Diophantos.
[O'Beirne, 1965, Chapter 12]
75. This is Yang Hui's solution.
[Needham, 1959, p. 60]
76. 'The ox leaves no trace in the last furrow, because he precedes
the plough. However many footprints he makes in the earth as he
goes forward, the cultivating plough destroys them all as it follows.
Thus no footprint is revealed in the last furrow.'
77. Alcuin answers: 'First there were 250 pigs bought with 100
shillings at the above mentioned rate, for five fifties are 250. On
division, each merchant had 125. One sold the poorer quality pigs at
three for a shilling; the other the better quality at two for a shilling.
The one who sold the poorer pigs received 40 shillings for 120 pigs;
The Solutions 203
the one who sold the better quality received 60 shillings for 120 pigs.
There then remained five of each sort of pig, from which they could
make a profit of 4 shillings and 2 pence.'
78. If the servant is not included in the count at each stage, then he
would arrive at the first manor having collected no men, so would
collect none there, and so on; the total collected would be zero!
Therefore the servant must include himself as the first soldier, and
the numbers on leaving each manor are 2, 4,8 ... , and on leaving the
thirtieth manor, 2
30
= 1,073,741,824.
79. They are cousins twice over, each having a parent who is sibling
to a parent of the other, in two ways.
80. 'To each son will come ten flasks as his portion. But divide them
as follows; give the first son the ten half-full flasks; then to the second
give five full and five empty flasks, and similarly to the third.'
81. Alcuin's solution, abbreviated, is: 1 cross with my sister, leave her
on the other side, and return. The other two sisters then cross, and
my sister brings the boat back. Then the other two men cross and one
returns with his own sister. Then he and 1 cross over, leaving our
sisters behind, and one of the women takes the boat back, and picks
up my sister who is carried over to us. Finally, the man whose sister
remains on the first bank, crosses over and brings her to us.
This takes a total of eleven crossings, which is more than necessary,
as the translator points out. A shorter solution is: 1 and my sister
cross, 1 return; the other women cross, my sister returns; 1 and my
sister cross again and 1 return; the other two men cross, and my sister
returns; 1 and my sister cross over. This is a total of nine crossings.
82. 'I would take the goat, and leave the wolf and the cabbage. Then
1 would return and take the wolf across ... and take the goat back
over; and having left that behind 1 would take the cabbage across; 1
would then row again and having picked up the goat take it over
once more. By this procedure, there would be some healthy rowing,
but without any lacerating catastrophe.'
83. 'First the two children get into the boat, and cross the river; one
of them brings the boat back. The mother crosses in the boat, and her
child brings the boat back. His brother joins him in the boat and they
go across, and again one of them takes the boat back to his father.
204 Penguin Book of Curious and Interesting Puzzles
The father crosses, and his son ... having boarded, returns to his
brother; and both cross again. With such ingenious rowing, the
sailing may be completed without shipwreck.'
84. This problem is related to earlier problems (see problem 37
above) based on Islamic and Roman law. The information given is
not sufficient, though it might be sufficient if we assume familiar
legal principles not stated in the problem.
Alcuin gives the mother the average of the two amounts she would
have received in the two cases; the average of t and n is t so the
mother receives 320 shillings. The son and the daughter each receive
one half of what they would have received if born alone, that is, 360
shillings and 280 shillings respectively.
The translator suggests adding the original fractions they expected,
i + n + t (the mother's average expectation) for a total of f. Now
multiplying their original expectations by i, the mother receives 432
shillings, the boy 336 shillings and the daughter 192 shillings.
85. Alcuin solves this much as Gauss solved, when a small boy in
school, the problem of quickly summing the integers from 1 to 100.
Alcuin explains: 'Take the one which sits on the first step, and add it
to the 99 which are on the 99th step, and this makes 100. Also the
second and the 98th, and find again 100. So for each step ... will
always give 100 between the two. The fiftieth step is on its own, not
having a pair, and similarly the 100th is on its own. Join altogether
and get 5050.'
86. There can be no integral solution. However, a rational solution
can be found from any three squares in arithmetical progression,
whose common difference is of the form 5p', on division by p.
Fibonacci found 31', 41' and 49', whose common difference is
720 = 5 X 12'. Hence his solution is 41112.
87. Let S be the original sum and 3x the sum returned equally to the
three men. Before each man received a third of the sum returned, they
possessed !S - x, tS - x and 1S - x respectively. Since these are the
sums that they possessed after putting back !, t and! of what they
had first taken, the amounts first taken were 2(S12 - x), 1(S13 - x)
and 1(S16 - x), and these amounts sum to S.
This gives the equation 7S = 47x, which is indeterminate, as is
inevitable from the original conditions, which only concern propor-
tions with no stated fixed amount.
The Solutions 205
Fibonacci chose the simplest values, S = 47 and x = 7. The sums
taken by the men from the original pile are then 33, 13 and 1.
[Eves, 1981, pp. 166-7]
88. Assuming that the rabbits are immortal, the number of new pairs
produced per month follows this sequence (Leonardo omitted the
first term, supposing that the first pair bred immediately):
1 2 3 5 8 13 21 34 55 89 144 233
This is the famous Fibonacci sequence, so named by Lucas in 1877.
Each term is the sum of the previous two terms. For many of its
wealth of properties, see the Penguin Dictionary of Curious and
Interesting Numbers, p. 61 et seq.
Binet proved in 1844 that the nth Fibonacci number is given by the
formula:
F = (1 + Js)" - (1 - Js)"
" 2" x Js
89. This is equivalent to a cistern problem. Instead of three pipes
pouring water into a pool at different rates, three animals remove
flesh from the sheep at different rates.
Fibonacci argues that in sixty hours (a conveniently chosen number)
the lion would eat fifteen sheep, the leopard would eat twelve, and
the bear ten, a total of thirty-seven. Therefore they will eat one sheep
in 60/37 = 1 ~ hours.
[Fauvel and Gray, 1987]
90. Suppose the last son received N bezants. The last-but-one son,
who also received N bezants in total, received (N - 1) + ~ of the
remaining bezants at that stage. There were therefore seven bezants
remaining after he had taken N - 1, so there were N + 6 bezants to
be distributed after the previous son had received his share, and these
N + 6 bezants make the two shares of N each taken by the last two
sons. Therefore N = 6 and, working backwards, the father left an
estate of thirty-six bezants, which was divided between six sons.
[Eves, 1976, p. 230]
91. He took 382 apples. The numbers left after he gives one half and
one apple more to successive guards, are (382), 190, 94,46,22, 10,4,
and 1 for himself.
[Eves, 1976, p. 231]
206 Penguin Book of Curious and Interesting Puzzles
92. In 63 day-plus-nights it moves upwards 63 x (j - t) = 147/5 =
29i.
The final i will be covered in i -;- i = n, of a day, after which the
serpent will not slip back. So the total time taken is 63n, days.
93. Draw a circle through the highest and lowest points of the statue,
so that it touches a horizontal line through the eye of the spectator.
The spectator should stand with his or her eye at that point.
There is some uncertainty in this solution, since the statue is not a
vertical rod! Turning the figure on its side, the solution for the rugby
conversion problem is the same. The conversion should be taken from T.
goal line
T
The Solutions 207
Because of 'the angle in the same segment is equal' property, the
same angle will be subtended by the goal-posts at any point on this
circle; any point outside it will subtend a smaller angle and any point
inside it, a larger angle. (Both true, though not part of the usual
statement of the theorem.) Choosing the circle to just touch the line
ensures that all other points of the line are outside the circle.
94. The speeds of the couriers are 250/7 and 250/9, so their speed of
approach to each other is 250 (1/7 + 1/9) = 250 x 16/63 and they
will meet in 63/16 = 3* days.
95. Place two nuns in each of the corner cells, leaving the middle cells
empty.
2 2
2 2
96. He worked for eighteen days and did not work on twelve days.
[Eves, 1976, p. 235]
97. Fill the 5 jar and fill the 3 jar from it, leaving 2 pints in the 5 jar.
Empty the 3 jar back into the cask and pour the 2 pints into the 3 jar.
Next, fill the 5 jar and fill the 3 jar from it, which takes 1 pint,
leaving 4 pints exactly in the 5 jar.
98. By working backwards, Josephus and his companion placed
themselves at positions 16 and 31 in the circle of 41 souls.
99. The Christians and Turks should be placed in the following
circular order, in which the first person follows the last, and the
counting starts with the first person: CCCCTTTTTCCTCCCTCT
TCCTTTCTTTCCTT.
208
100.
Penguin B.ook of Curious and Interesting Puzzles
• Full
o Empty
o
130
90
101. As Dudeney pointed out, the central square can be reached by
none of the knights, and the other eight squares can be reached from
just two other squares, as illustrated by the circular figure.
From the circular figure, It is trivially easy to read off the moves
that must be made: any piece may make the first move, in either
direction, and the other pieces then chug round the circle in the same
direction.
The Solutions 209
102. Tartaglia gives a pheasant to Piero, and to Piero's son Andrea,
and to Andrea's son Filippo.
103. Previous authors had assumed that such conditions meant
that the inheritance had to be divided literally in the proportions
given. It therefore made no difference whether the given fractions
added up to less than one, to one exactly, or to more than one.
Tartaglia suggested the device of borrowing an extra horse, making
eighteen horses; each person can now be given a whole number of
horses, totalling seventeen in all, and the borrowed horse returned to
its owner.
A solution by this precise trick method - by borrowing exactly one
horse and then returning it - only exists for certain conditions of the
problem. If n horses are to be divided among three sons, who receive
respectively lIa, lib and lie, then the seven possible values of n,a,b,e
are: 7,2,4,8; 11,2,4,6; 11,2,3,12; 17,2,3,9; 19,2,4,5; 23,2,3,8; 41,2,3,7.
[Gardner, 1978a)
However, if you do not mind whether you borrow-and-return or
lend-and-recover, and if you do not mind how many horses are
borrowed or lent, then the method always works whenever the
numerator of the sum of the fractions is equal to the number of
horses to be divided. For example, suppose you wish to share 31
horses in the ratios 112 to 113 to 115 (sum 31130). Lend 1 horse, take
112, 113 and 115 of 30, which is 15, 10 and 6, and then take your
borrowed horse back. For 112, 113, 114 and 115, totalling 77/60, share
77 horses by lending 17 and taking them back!
104. Let the barrel originally contain x pints of wine. After one
removal and replacement, its strength will be (x - 3)/x, and the
amount of wine removed on the second removal will actually only be
3(x - 3)/x, and on the third removal, the wine removed will be
3(x - 3 - 3(x - 3)/x)
x
The total wine removed is one half the original quantity, and the
equation simplifies neatly to 2(x - 3)3 = xl, or x = (3 X 2'/3)/
(2'/3 - 1) = 14.54 pints.
105. If the answer given to you is N, then the number originally
chosen is 2N or 2N + 1, depending on whether you were told at the
second step that the answer was even or odd, respectively.
210 Penguin Book of Curious and Interesting Puzzles
106. If the remainders on dividing by 3, 4 and 5, are A, Band C
respectively, then the original number is the remainder when 40A +
45B + 36C is divided by 60.
Suppose that the number chosen is x, so that the three equations
are x = 3a + A = 4b + B = 4c + C. Then, multiplying the equa-
tions by 40, 45 and 36 respectively:
40x = 120a + 40A
45x = 180b + 45B
36x = 180c + 36C
Adding, 121x = a multiple of 60 + (40A + 45B + 36C). Therefore x
and 40A + 45B + 36C have the same remainder when divided by 60,
and since x was chosen to be less than 60, it equals that remainder.
107. Twenty counters. The numbers of counters in the hands of the
first and second person are, in three stages: x and 3x; x - 5 and
3x +5; (x - 5) + 3(x - 5) and 3x + 5 - 3(x - 5), and the last
number is always 20.
108. Tartaglia (and Fibonacci before him) had considered the problem
of the weights required, if they can only be placed on one side of the
balance, and concluded that the best solution has weights in the
sequence of powers of 2: 1, 2, 4, 8, 16, 32, and so on. This is the same
as saying that each integer can be represented uniquely in the binary
notation.
Bachet gave the solution 1, 3, 9 and 27 when both pans may be
used. The basic idea is that every number is one more or one less than
a multiple of 3. Thus 32 = (3 XII) - 1 = 3(3 x 4 - 1) - 1 =
3(3 x (3 + 1) - 1) - 1 =3' + 3' - 3 - 1. Therefore 32 Ibs can be
weighed by placing the 27 and 9 Ib weights in one pan and the 3 and
lIb weights in the other.
It is plausible that Bachet's solution is in some sense best possible,
merely because it is so simple and elegant. This was proved in 1886
by Major MacMahon, who used the method of generating functions
discovered by Euler to show that there are eight possible sets of
weights, apart from the one-scale solution, 1,2,4 ... 32. Denoting the
number of each weight by a superscript, they are:
1
40
; 1,313; 1\ 9
4
; 1,3,9
4
; P\ 27; 1,3
4
,27; 1\ 9, 27; 1,3,9,27.
Thus Bachet's solution does indeed use the fewest weights and is also
the only solution in which all the weights are different.
The Solutions 211
109. The easiest way to construct Heronian triangles is to fit two
right-angled triangles together.
5 9
Here are a 5-12-13 and 9-12-15 triangle fitted together. The resulting
triangle has altitude and sides 12-13-14-15, and is the only possible
such triangle. The area is 84.
110. The right-angled triangles 5-12-13 and 6-8-10 each have area
equal to perimeter. The three proper Heronian triangles with this
property are 6-25-29; 7-15-20; 9-10-17.
111. In the solution to 109, place the same two triangles so that they
overlap. The obtuse-angled triangle with sides 4-13-15 has area
54 - 30 = 24.
Proofs of these results are not so simple; one method is to write
Heron's formula in the form
which can be written in the form
(4A)' + (b
'
+ c
'
- a
'
)' = (2bc)'
This is of the form p' + q' = r and has parametric solutions, p =
m
'
- nl, q = 2mn and r = m
'
+ nl.
212 Penguin Book of Curious and Interesting Puzzles
112. Each knife blade goes under the blade of one other knife and
over the blade of the third. So arranged, they can easily support a
glass of water well above the table surface.
113. Force the tips of three knives into the stick, so that the knives
hang well below the finger. The centre of gravity of the entire
arrangement will then be below the finger tip and will be stable.
114.
115. Bend the straw and insert into the neck of the bottle, which can
then be lifted.
The Solutions 213
116.
Van Etten gives variants of this puzzle, in which the combination
of holes to be plugged are different, such as square, circular and oval.
The principle is the same.
117. He is standing at the centre of the earth.
214 Penguin Book of Curious and Interesting Puzzles
118. Van Etten gives the same solution - they are at the centre of the
earth, and ascending in opposite directions. However, it is also true
that if they ascended two vertical ladders on the earth's surface, they
would also be moving apart, albeit by a minuscule amount.
119. When he is standing at the North Pole.
120. Place the point on the surface of a sphere and draw a circle,
which will be smaller than the circle drawn by the same compass on a
plane surface. Alternatively, place the point of the compass at the
apex of a circular cone, and draw a circle on its surface.
121. Wrap the paper on which you are to draw the oval round a
cylinder. The compass will then draw an oval.
122. One horse travelled east and the other travelled west, the first
gaining in the number of days it lived, and the second losing.
123. First they sold their apples at 1 penny each, then later in the day
they sold them at 3p each. A sold 2 @ Ip and 18 @ 3p, making 56
peace. B sold 17 @ 1 P and 13 @ 3 p, and C sold 32 @ 1 p and 8 @
3p, each also making 56 pence.
124. The number of individuals in the world far exceeds the number
of hairs on the head of anyone of them. Therefore if you start to pick
out individuals with given numbers of hairs on their heads, you will
be forced to pick an individual with a number of hairs that you have
already counted once, long before the population of the world is ex-
hausted.
This is the first known example of the 'pigeonhole principle',
which says that if you have N + 1 objects to place in N pigeon holes,
then one hole must contain at least two objects.
125. The sum of the distances will be a minimum when the lines OA,
08 and OC all meet at 120°. (This is a general principle that applies
to all such minimum networks of 'roads'. If one of the angles is
greater than or equal to 120°, the point sought is at that vertex.)
To construct point 0 with ruler and compasses, draw equilateral
triangles outwards on each side of the triangle, and draw the circumcir-
cles of each new triangle. These circles will pass through a common
point, which is O. This construction works, because of the property
that the opposite angles of a cyclic quadrilateral sum to 180°. Choos-
The Solutions 215
ing the extra triangles to be equilateral, with angles of 60°, ensures
that the opposite angles will all be 120° as desired.
It also happens that if the outer vertices of the equilateral triangle
are joined to the opposite vertices of the original triangle, they will all
pass through the point O.
[For further discussion see Honsberger, 1973, p. 24]
126. It is only necessary to 'tie' an ordinary knot in the strip and
carefully flatten it.
216 Penguin Book of Curious and Interesting Puzzles
127. Amazingly, the answer is that a cube of side slightly under
3j2
4
or approximately 1.060660, can be passed through a given cube of
side 1 unit.
....
. ........... """ .
.
.
.
.
~
. .
.. ...
•
In this figure, the square hole cuts the top face along the lines
EFGH, the bottom face along ABCD and the two vertical edges at X
and Y, as indicated by the dotted lines.
128. Typically, Newton expresses the problem in general form, rather
than giving values to the letters.
Suppose that each field contains initially the same amount of grass,
M, and the daily growth in each field is also the same, m, and that
each cow consumes the same amount of grass per day, Q.
Then
and
and
bM + cbm - caQ = 0
b'M + c'b'm - c'a'Q = 0
b"M + c"b"m - c"a"Q = 0
By a standard theorem in determinants, given that M, m and q are
not all zero, this determinant is zero:
b bc ca
b' b'c' c'a'
b" b"c" c"a"
The Solutions 217
Without usmg determinants, M, m and Q can be eliminated 'by
hand', to give
b"ee'(ab' - ba') + e"b"(be'a' - b'ea) + e"a"bb'(e - e') = 0
[After Dorrie, 1965, p. 9]
129. It might seem that the chances are equal, because the proportion
of sixes required to the number of dice thrown is constant. This is not
so.
'The chance of getting 1 six and 5 other outcomes in a particular
order is mW
5
• We need to multiply by the number of orders for 1 six
and 5 non-sixes. Therefore the probability of exactly 1 six is

Similarly, the probability of exactly x sixes when 6 dice are thrown is
x = 0,1,2,3,4,5,6
The probability of x sixes for n dice is
x = 0,1, ... , n
This formula gives the terms of what is called a binomial distribu-
tion.
'The probability of 1 or more sixes with 6 dice is the complement
of the probability of 0 sixes:
1 - 0.665
'When 6n dice are rolled, the probability of n or more sixes is
I 1 _ 'i
l

x=" x 6 6 <=0 X 6 6
Unfortunately, Newton had to work the probabilities out by hand,
but we can use the Tables of the Cumulative Binomial Distribution,
Harvard University Press, 1955. Fortunately, this table gives the
cumulative binomial for various values of p (the probability of
success on a single trial), and one of the tabled values is p = i. Our
218 Penguin Book of Curious and Interesting Puzzles
short table shows the probabilities, rounded to three decimals, of
obtaining the mean number or more sixes when 6n dice are tossed.
6n n P{n or more sixes)
6 1 0.665
12 2 0.619
18 3 0.597
24 4 0.584
30 5 0.576
96 6 0.542
600 100 0.517
900 150 0.514
Clearly Pepys will do better with the 6-dice wager than with 12 or 18.
When he found that out, he decided to welch on his original bet.'
[Mosteller, 1987, problem 19]
130. Bernoulli posed the problem in terms on n letters wrongly
placed into n envelopes, but the principle is the same. The general
formula for the number or ways of misplacing all the letters is:
(
1 1 1 1 -1")
n! 2i - 3! + 4! - 5! + . . . + --;;;-
When n = 7, the value is 1854.
[A complete solution is in Dorrie, 1965, p. 19]
131. One! This answer is independent of the number of letters and
envelopes.
[Newman, 1982, p. 22]
132. These two solutions have the features of simplicity and sym-
metry.
[Fisher, 1973]
The Solutions 219
133. No, it is not. The map of the river can be represented schemati-
cally like this:
The problem is now to trace out this figure with a pencil, passing
over every line once, and no line twice, without lifting your pencil
from the paper. This is only possible if the figure to be traced
contains either no vertices at which an odd number of edges meet (in
which case you may start at any point you choose, and trace the
figure so as to return to your starting point), or just two such
vertices, in which case you can only trace the figure by starting at one
and ending at the other.
The reasoning IS simple: in arriving at a vertex and then leaving it,
two of the edges meeting at the vertex are 'used up'. Therefore, any
vertex at which an odd number of edges meet (all of which must be
traversed) can only be a starting or an ending vertex, of which there
can be at most two.
All four vertices in the figure for the Bridges of Konigsberg are
'odd', and so the figure cannot be traversed.
134. Every edge has two ends, so the total number of edge-ends IS
even. But the number of edge-ends is also the total of the number of
edges meeting at each of the individual vertices, which must therefore
include an even number of odd vertices, since an odd number of odd
vertices would give an odd grand total.
This was one of the points established by Euler in his original
paper.
135. This is one solution. Typically the same letters are knight's
moves apart from each other, and the patterns formed by the shading
are similar.
220 Penguin Book of Curious and Interesting Puzzles
A B C 0 E
C D E A B
E A B C 0
B C D E A
0 E A B C
136. 19,013 years, 144 days, 5 hours and 55 minutes.
137. Sixty-six years.
138. Forty-five and fifteen years.
139. The given solution is something of a cheat, quickly referring to
one of Dr Hutton's textbooks. (Readers also occasionally stooped to
taking questions from published sources. As the editor remarks of
Question 97, 'It is evident that this question is composed from that in
page 225 of Ward's MathematicIan's Guide'!)
Assuming that the cylinder to be cut out has its base on the base of
the cone, it remains only to determine its height, as a proportion.
The volume of the cylinder is proportional to DF x PQ, and we
know that DElAP is constant. Therefore, it is required to maximize
AP' x PQ. In other words, given any line AQ, find a point on it, P,
such that AP' x PQ is a maximum.
For clarity, call AP x, and let the height AQ = L. Then we have to
maximize xZ(L - x), which is the same as maximizing xZ(2L - 2x).
But the latter is the product of three factors whose sum is constant, at
2L, and it will achieve its maximum value if all three factors are
equal. Therefore the maximum is when x = 2L - 2x and x = 2L13.
So AP must be two-thirds of AQ.
The Solutions 221
A
140. 2520 = 5 x 7 x 8 x 9.
141. This is the solution by Mr J. Hill: 'Call the number of hogs any
[one] woman bought x; the number her husband bought x + n;
money laid out by the woman is xx shillings; money laid out by the
husband is xx + 2nx + nn shillings. Equation
xx + 2nx + nn = xx + 63
[Therefore] x = (63 - nn)/2n.
If n = 1, then x = 31 and x + n = 32; hence some woman bought
31 hogs, and her husband 32. If n = 3, then x = 9, and x + n = 12;
therefore some other woman bought 9, and her husband 12. If n = 7,
then n + x = 8; [therefore] some woman bought 1, and her husband
8. Consequently,
Hendrick bought 32 and his wife Anna 31
CIa as . . . . . . . . . .. 12 ........... Catriin 9
Cornelius. . . . . . . 8 ........... Geertrick l'
222 Penguin Book of Curious and Interesting Puzzles
As another solver, Mr N. Farrer, noted, the numbers of hogs
bought by husband and wife are such that the differences of their
squares are 63, which gives three possible pairs only: 8,1; 12,9; 32,31.
142. 5rl:- minutes past 7 o'clock, or approximately 7.05 and 27 seconds.
143. The extra square has actually become the long and very thin
parallelogram formed by the 'diagonals' of the second figure, as
exaggerated here.
8 5
5 5
5 8
This can be checked by calculating the slopes of the different
portions of this 'diagonal', which are 5/13 and (8-5)/8 = 3/8. These
fractions are close, but not equal.
The fractions are so close because they have been conveniently
taken from the Fibonacci sequence, 1 1 2 3 5 8 13 21 34 55 ... which
has the property that of any four consecutive terms, the products of
the outer pair and of the inner pair differ by one. By taking four
consecutive terms later in the sequence, the paradox becomes even
more difficult to detect by eyesight alone.
The Solutions 223
144.
[Delft and Bottermans, 1978)
145. The rowers move at 4 miles per hour relative to the water, so
they approach each other at 8 miles per hour, and will close the 18
miles between them in 21 hours. If the water were still, their meeting
would be at the mid-point, 9 miles from each town. But due to the
current, their meeting place will have moved at H miles per hour,
over the 21 hours, downstream from Haverhill towards Newburyport,
a total drift of 3 ~ miles. So they meet 5i miles from Newburyport.
146. 9 9 ~ .
147. SIX
From IX take
XL
IX S
X leaving I
L X
148. 2 + 4 + 6 + 0.8 = 12.8 and 1 + 3 + 7 + 9/5 = 12.8.
149. One-third of TWELVE is LV = 55 in Roman numerals; one-
fifth of SEVEN is V = 5, and 55/5 = 11.
150. 123456789 x 8 = 987654312.
224 Penguin Book of Curious and Interesting Puzzles
151. This is jackson's solution, arranged in tabular form:
barrel first container second container
12
7 5
7 5
2 5 5
2 3 7
9 3
9 3
4 5 3
4 7
11
11
6 5 1
6 6
152. If cells are filled with the numbers (reading left to right, top to
bottom) 7, 0, 5; 2, 4, 6; 3, 8, 1, then the square is magic in the usual
way, by addition of the rows and columns. If each number is replaced
by the matching power of 2, then it is magic by multiplication of
rows and columns. So, one solution, reading the rows from left to
right, top to bottom, is: 128, 1,32; 4, 16, 64; 8, 256, 2.
153. This is jackson's solution:
4 9 5 16
15 6 10 3
14 7 1 1 2
1 12 8 13
The Solutions 225
154. Jackson merely states that the true weight is the mean propor-
tional (or geometric mean), that is, the square root of 16 x 9 or
121hs.
Suppose that the long and short arms of the balance are of length p
and q respectively, and the true weight is W. Then Wp = 16q and
Wq = 9p, from which W'pq = (16 x 9)pq and the conclusion
follows.
155. A shoe.
156. This is jackson's solution. It generalizes to dividing a circle into N
parts by N lines of equal length. Divide a diameter into N equal parts,
and construct a sequence of semi-circles on either side, following this
pattern. The tadpole-shaped regions at each end and the N - 2 snake-
like regions between will all have equal area, and will be bounded pairs
of lines each equal in length to half the perimeter of the circle.
157. Draw it on a sphere, taking, for example, one of the poles and
any two points on the equator which are separated by one q u ~ r t e r of
the earth's circumference (taking the earth to be spherical).
158. 0 is the centre of the circle. With the compass open to the
radius of the circle, mark off the points C, X, and B in succession.
Then with radius BC, and centres A and B, draw arcs to intersect at
D.
Then DO is the length of the side of the required square. Marking
AE so that AE = DO, and similarly marking F, produces the square.
226 Penguin Book of Curious and Interesting Puzzles
D
A B
F
Also, since AOE = 90° and AOC = 60°, COE = 30° and CE is one
side of an inscribed dodecagon.
159. 'Suppose one place to lie directly under either of the poles, a
second 10 degrees on this side, and a third 20 degrees on the other,
under the same meridian circle, then they will all differ in latitude,
and likewise in longitude, since the pole contains all degrees of longi-
tude.'
The figures chosen by Jackson are, of course, quite arbitrary.
160. The South Pole.
161. The reference to Naples and the situation of the village in a low
valley are mere Rim-Ram, worthy of Sam Loyd. The fact is that any
place on earth, the poles apart, varies daily in distance from the sun
because of the earth's rotation, being a maximum for places on the
equator, and a minimum of zero for the actual poles. 3000 miles is
jackson's estimate of the variation for Naples, based on the earth
having a radius of about 4000 miles.
162. The island is Guernsey. (Any of the other Channel Islands
would do.) Guernsey is 26 miles from France, but England, from
Dover to Calais, is only 21 miles from France.
The Solutions 227
163. The Christian sets off from the Jew's abode, travelling East, and
the Turk does likewise, but travels West. When they meet again at
the Jew's, by their own reckoning, having respectively lost a day and
gained a day while travelling round the world, they will each be able
to celebrate their own sabbath on the same day in the same place!
164. The traveller's journey has been right round the world. His head
is about 6 feet from the ground, and so the radius of the giant circle
travelled by his head is about 6 feet greater than that of the circle
travelled by his feet. This difference in radius produces a difference in
circumferences of about 27t x 6 feet, or about 36 feet = 12 yards.
Jackson attributes this idea to Whiston's commentary to his edition
of Euclid.
165. At In, 2A-, 3rl- ... hours.
166. There are fourteen different arrangements, ignoring colours,
each of which can be coloured in two ways, making a total of twenty-
eight. (Some pairs of these arrangements would be equivalent if the
tiles were painted on both sides, and arrangements of tiles could be
turned over.)
l/1SJ21 NZ1SJ
rsrsN l/1/1/1
tsN/1 l/1/lSJ
~ ~ ~ 6 ¥
~ ~ ~ 6 ¥
228 Penguin Book of Curious and Interesting Puzzles
167. A: 14¥, B: 17* C: 23f.-
168. This solution is equivalent to the problem of dissecting a Greek
Cross into a square: assemble the five squares to form the cross, and
the solution appears immediately.
This dissection of a Greek Cross is one of an infinite number which
depend on the possibility of tessellating the plane with identical
crosses:
Take any four corresponding pomts to form the vertices of a
square, equal in area to any Cross, and the square at once forms a
dissection. Since the four corresponding points can be chosen in an
infinite number of ways, there are an infinite number of solutions.
The Solutions 229
169. The two smaller squares in the Pythagoras figure will tessellate
also, and by joining corresponding points together, an infinite number
of dissections of the smaller squares into the larger are found. In
every case, the pieces require only to be slid, without rotation, into
their new positions.
170.
r - - - - - - - ~ / ~ - - - - - , - - - - - - --- --;7
/ /
/ /
/ /
/ /
/ /
/ /
/ /
/ /
/ /
/ /
/ /
/ /
/ /
/ /
~ / _ - - - - - - - - - - - - ~ /
,
,
,
,
,
,
,
,
,
,
,
,
,
,
, /
, /
" //
, /
, /
, /
, /
, /
, /
, /
, /
, /
, /
V
230 Penguin Book of Curious and Interesting Puzzles
171. Let the radius of the base be R and the height H. Then the total
surface area is nR
2
+ 2nRH, and the volume, which is fixed, is V =
nR2H.
Therefore the surface area is n(R 2 + ~ + ~ ) .
nR nR
This is the sum of three terms whose product is constant. It will
therefore be a minimum if all three terms are equal, that is, when
~ = R 2 and R = 3 I ~ n .
nR' ~ .
172. The bottles must be mixed in the proportion of 3( = 8 - 5) of
the first to 2( = 10 - 8) of the second, that is, 3 bottles of the first
costing 30s and 2 bottles of the second, costing lOs; total, 5 bottles
costing 40s, or 8s a bottle.
173. The largest possible rectangle has its base on one side, and its
top edge joins the mid-points of the other two sides. Its area is then
one half of the area of the triangle, and therefore it makes no
difference which is chosen to be the base of the rectangle: three
different maximum-area rectangles have equal areas.
174. Imagine that equal weights are placed at the vertices of the
original polygon. Then replacing these equal weights by an identical
set, placed at the mid-points of the sides, will not change the centre
of gravity of the arrangement. Therefore the centre of gravity is
unchanged by any number of repetitions of the process, and the
sequence of polygons contracts to a point that is the centre of gravity
of equal weights at the original vertices of the polygon.
175. Draw a line through the given point, X, so that PX = XQ.
Imagine that the line is rotated very slightly about X to cut the two
lines at P' and Q'. Then the triangles PXP' and QXQ' will have equal
areas, to a first approximation. The position PXQ is therefore a
limiting point for the area of the enclosed triangle, and since it cannot
be a maximum, it must be a minimum. This intuitive answer is
confirmed by calculation.
The Solutions 231
176. Ozanam gives the solution square in this algebraic form, so that
any numbers can be substituted for a, band c.
a
2ac
c
-
a+c
2ab 2bc 2abc
a+b b+c 2ab+ac-bc
b
2abc abc
2ac+ab-bc ab+ac-bc
To get whole-number solutions, he chooses values of a, band c to
give this square:
1260 840 630
504 420 360
315 280 252
232 Penguin Book of Curious and Interesting Puzzles
177. The secret is to make the sum, after one of your turns, equal to
a number in the sequence 1, 12, 23, 34, 45, 56, 67, 78, 89, 100. Once
you have achieved this (which is easy enough if you start the game -
you just choose 1 as your first call), then you can keep to the
sequence by calling out the difference between 11 and your opponent's
last call.
If your opponent starts the game, and knows the trick, then of
course you must lose, but only alternate games!
178. Every solution can be rotated and reflected to reproduce seven
other solutions, which, however, will in some cases be identical to the
original solution due to its symmetries.
There are twelve basic solutions on the full chessboard. Each can
be described by a single 8-digit number, by reading off the position of
each queen in each column, starting from one end. With this notation,
the twelve solutions are: 41582736; 41586372; 42586137; 42736815;
42736851; 42751863; 42857136; 42861357; 46152837; 46827135;
47526138; 48157263. [Rouse Ball, 1974, p. 1711
179. Using the same notation as solution 178, there is one basIC
solution on a 4 x 4 board: 3142. There are twO basic solutions on a
5 x 5: 14253 and 25314; and one on a 6 x 6: 246135.
On a 7 x 7 board there are six basic solutions.
The Solutions 233
180. This is the commonest solution offered:
It is a special case of Pappus's theorem, which says that if points A, B
and C are taken on one line, and A', B' and C' taken on another, and
joined as in the figure, then the points P, Q and R will lie on a line. (P
is the meet of BC' and B'C, and so on.) If it happens that B, Q and B'
also lie on a line, then the figure satisfies the conditions of the
problem.
A more general solution is any figure for Desargues's theorem. In
other words, take any two triangles, such that the lines through
matching vertices meet in a point P.
p
The pairs of corresponding sides will then also meet in pairs on a
line L.
234 Penguin Book of Curious and Interesting Puzzles
181.
* * '*
* * *
* * *
* * *
0
* * *
* * *
* * *
* * *
182. Plant three of the trees at the base of a steep mound, at equal
distance from each other, and plant the fourth tree on top, at equal
distance from the other three.
183. This solution (ignoring the dotted lines) can be varied, by
varying the triangles. Indeed, any two overlapping triangles will do.
184. The four parts are 8, 12, 5 and 20.
185. 'Place 4 on 7, 6 on 2, 5 on 8, and 3 on 1. Recollect always to
begin with either 4 or 5. The same trick may be thus performed: place
5 on 2, 3 on 7, 8 on 6, and 4 on 1.'
The Solutions 235
l ~ · N N E
HUT
187.
- ~ - - - j - - - I - - -
VII
-><------1-+--
VIII
188. This puzzle was the basis for one later made famous by Sam
Loyd.
236 Penguin Book of Curious and Interesting Puzzles
189.
190.
-
I
I
r--
-
191. This problem has reappeared many times, often in its simplest
guise - an L-shape formed from three-quarters of a square, and the
demand that it be dissected into four identical parts.
The Solutions 237
192. Fold one edge of the square on to its opposite edge to get a
middle line. With the middle line horizontal, fold one lower corner on
to the middle line, so that the fold passes through the other lower
corner. Repeat, using the other lower corner. These two folds and
one edge form an equilateral triangle.
[Tom Tit', n.d.]
193. 'Roll the paper into a short compact roll. Make two parallel
cuts across the roll, each being about one half an inch from the other
end. Then make a long cut parallel to the axis of the roll and
terminated by the cross cuts. This will produce a gap in the roll.
Holding the roll lightly in the fingers ease out the ends of the first
strip, which lies at the bottom of the gap, then, taking the strip with
the teeth and holding the roll lightly by its two ends, slowly draw the
strip out of the gap ... the whole inside of the roll will be drawn
through the gap, the connecting parts of the successive strips being
twisted. The final result will be a series of paper strips which serve as
rungs of a ladder, whose upright sides are formed by the twisted
parts.'
194. Trace round a shilling and cut out the circle to make a circular
hole. This hole cannot be increased in size as long as the paper is Rat,
without tearing the paper. However, if the paper is folded across a
238 Penguin Book of Curious and Interesting Puzzles
diameter of the circular hole, and a larger coin - the half-crown was
just the right size - is placed within the fold next to the hole, then by
bending the paper without tearing, the hole can be enlarged suffi-
ciently to allow the larger coin through. The maximum diameter of
the hole when the paper is bent is fl, or very slightly over 1.4, times
its original diameter.
195. Draw a rectangle inside the given sheet so that there IS a
sufficient margin for gluing the resulting envelope.
Bisect the shorter sides, and draw a circle, centre at the centre of
the rectangle, passing through A and B, to cut the other sides at a pair
of opposite points, C and D. Join ABCD. This is the front of the
envelope, which is just covered completely at the back, when the four
outer triangles are folded inwards.
196. Like this (the principle can be used to cut any board in the same
or related proportions):
2
I
4
2
4
The Solutions 239
197.
0 0 0
0
0 0
0
0 0
0 0 0
198. Fold the card in half, and make cuts with scissors as shown.
Finally cut down the original central fold, omitting the two end
portions. The card is now reduced to a strip which may be opened
out and passed over a person.
199. Number the coins from 1 to 10, in sequence. Place 4 on 1,7 on
3,5 on 9, 2 on 6, and then 8 on 10.
240 Penguin Book of Curious and Interesting Puzzles
200. Here are a handful of solutions:
123 - 4 - 5 - 6 - 7 + 8 - 9 + 0 = 100
It + 98* + 0 = 100
80* + = 100
70 + 24-& + 5t = 100

201. Giraffe, lion, camel, elephant, hog, horse, bear, hound.
202.
203. Four strokes.
204. Scratch the table cloth and the 20p coin will emerge from under
the edge of the tumbler.
205.
ONE
206.
LOVE
The Solutions 241
207. Before picking up the handkerchief, fold your arms. Then pick
up the handkerchief, and unfold your arms. The 'knot' which was in
the folding of your arms is transferred, as it were, to the handker-
chief.
208. Fold the piece of paper on which the sum is written, like this, so
as to obscure the figure 300, and so that the other two lines form the
new number 707. The new sum has the same total as the old.
3/8
701
209. Jump 9 over 4, 5, 7, and 1; 3 over 2; 6 over 8 and 3; and 9 over
6.
210. Arrange the matches to form the edges of a regular tetrahedron.
211.
o o o
o o o o
o 0
242 Penguin Book of Curious and Interesting Puzzles
212.
213. Jane, Ann, Joe, Bet, Rose and Jim earn, respectively, 3s 2d,
2s 7d, 15 11d, 15 5d, 1s 1d and 8d per week.
214. The number of letters contained in each numeral word.
215. The four figures are 8888, which on being divided horizontally
along the middle line become a row of zeros, or nothing.
216. (a) 19!
(b) Place one of the coins on the table, then keeping the hands
apart, take it up with the other hand.
(c) Draw it round his body.
(d) 8!.
217. The squirrel takes out each day one ear of corn and his own two
ears.
The Solutions 243
218.
219. The difference in their ages.
221. 35170 + 148/296 = ! + t = 1
222. The original puzzle states that the two digits transferred are 28.
The answer is 285714, which is the period of the decimal fraction t =
0.i85714.
This period has the property that any circular rearrangement of the
digits is a multiple of the period of t, 142857.
223. The reply of most people is, almost invariably, that the hatter
lost £3 19s Od and the value of the hat, but a little consideration will
show that this is incorrect. His actual loss was £3 19s Od less his trade
profit on the hat; the nett value of the hat, plus such trade profit,
244 Penguin Book of Curious and Interesting Puzzles
being balanced by the difference, 21s, which he retained out of the
proceeds of the note.'
224.
.

, .' .' .

•
225. Remove the dotted matches, and three small triangles, three
medium-sized and the outer triangle are left, a total of seven triangles
remaining. In the original figure there are a total of thirteen triangles.
/\

226. The old gentleman was a Widower with a daughter and sister.
The old gentleman and his father (who was also a widower) married
two sisters (the wife of the old gentleman havmg a daughter by a
former husband); the old gentleman thus became his father's brother-
in-law. The old gentleman's brother married the old gentleman's
step-daughter; thus the old gentleman became his brother's father-in-
law. The old gentleman's father-in-law married the old gentleman'S
sister, and the old gentleman thus became his father-in-Iaw's brother-
in-law. The old gentleman's brother-in-law married the old gentle-
The Solutions 245
man's daughter, whereby the old gentleman became his brother-in-
law's father-in-law.
227. 864 - 72 = 792.
228. One person received his herring on the dish.
229. There is essentially only one solution, when the symmetries of
the dodecahedron are taken into account, which is easily read off the
plane map. For example, go round the minor pentagon, move to the
'star' pentagon and go round it, and then end by going round the
outer pentagon.
230. However the dogs run, the distance between each dog and the
dog he is chasing will be reduced from the initial 100 yards at the rate
of 3 yards per second. They will therefore meet in 33! seconds, and by
symmetry they will collide at the centre of the field.
231. De Morgan was born in 1806, and so was 43 in the year 1849 =
4Y.
232. This is a minimum solution, In forty-six moves, due to Dudeney.
Note that it is generally only necessary to name the piece that moves
at each turn. The '*' symbol indicates that a piece has jumped,
leaving the central cell empty:
Hhg*Ffc"CBHh*GDFfehbag"GABHEFfdg"Hhbc"CFf"GHh*
233. The maximum value which cannot be made is 63. Higher
numbers can always be made; thus, 64 = 2 @ 17 + 6 @ 5.
234. Yes. Move B one square to the right and move A round the
circuit to the right of B. Interchange C and D by moving round the
shaded cells and then shunt B-A to the left.
235. First, train B advances, and backs its rear half into the cul-de-
sac, uncouples it, and moves well forward of the junction. Second, A
passes the junction, backs, and joins to the rear half of B, which it
then draws out of the cul-de-sac and backs to the left. Third, the
front half of train B backs into the cul-de-sac. Fourth, train A
uncouples the rear half of train B and proceeds on its way. Train B
can now leave the cul-de-sac and join its rear half and proceed.
246 Penguin Book of Curious and Interesting Puzzles
236. No. The planks can be used together, provided that their length
is at least
2J2W
3
where W is the width of the moat (with a little allowance for the
planks to overlap each other and the bank).
x
w
ff
2J2 x 10
Since 8 < the given planks will not suffice.
3
237. The solution can be represented visually, like this:
New
o 1234567891011121314151617 days
The Solutions 247
The line AB represents the ship leaving Le Havre today. It passes
thirteen ships at sea, and meets two more ships, one in each harbour,
a total of fifteen.
[Attributed to Lucas in Kordemsky, 1972, problem 255]
238. The principles are the same whatever the number of discs.
Suppose, therefore, for simplicity, that there are eight discs to be
moved, from peg A to peg B. Number the discs 1 to 8 from the top
downwards. The simplest rules are: .
1. Always move an odd numbered disc, on its {irst move from peg A,
to peg C, and an even numbered disc, on its {irst move from peg
A, to peg B.
2. Move disc 1 every second move, disc 2 every fourth move, disc 3
every eighth move, and so on.
Following this rule, the sequence for eight discs will start: l-C, 2-B,
I-B (the first two discs have now been transferred to peg B, solving
the problem for just two discs); 3-C, I-A, 2-C, l-C (leaving discs 1
to 3 on C); 4-B, I-B, 2-A, I-A, 3-B, I-C, 2-B, I-B (leaving the first
four discs on peg B).
Notice that disc 1 visits the pegs repeatedly in the order C-B-A-C-
B-A ... ; disc 2 visits them in the order B-C-A-B-C-A so that its
visits 'rotate' in the opposite direction, and similarly for the remaining
discs.
The next move is to place disc 5 on to peg C and repeat the process
so far, to leave all discs up to 5 on C. Then place disc 6 on Band
repeat to get all discs up to 6 on B, place disc 7 on C and repeat to get
all discs up to 7 on C, and finally put disc 8 on peg B and repeat the
entire process to transfer the seven smaller discs from peg C to peg B.
The number of moves taken to move n + 1 discs is one more than
twice the number needed to move n discs. It is therefore 2" - 1. In
Lucas's original story the number of moves required is therefore
2
64
- 1, which at one move every second amounts to more than
500,000,000,000 years.
239. The probability is i. The following explanation is due to Howard
Ellis, one of Gardner's readers. Let Band W' stand for the black or
white counter which is in the bag at the start and W' for the added
white counter. After removing a white counter there are three equally
likely states:
248 Penguin Book of Curious and Interesting Puzzles
In bag
WI
W'
B
Outside bag
W'
WI
W'
In two out of three cases, a white counter remains in the bag.
[Carroll, 1958, problem 5, and Gardner, 1981, p. 189]
240. This is equivalent to asking for the relative volumes of a regular
tetrahedron and a regular octahedron. Carroll solved this problem by
calculation, but it is solved more efficiently by visualization. Fit two
such pyramids together to make a regular octahedron and inscribe it
in a regular tetrahedron.
The complete tetrahedron has 2 x 2 x 2 = 8 times the volume of
any of the four small tetrahedra affixed to alternate faces of the
o'ctahedron. Those four tetrahedra therefore occupy in total one half
of the volume of the large tetrahedron, and the octahedron occupies
the other half.
The pyramid which is one half of the octahedron therefore occupies
one quarter of the complete tetrahedron and is equal in volume to
two of the small tetrahedra.
[Carroll, 1958, problem 49]
241. This is Carroll's own solution:
'It may be assumed that the three points form a triangle, the chance
of theIr lying in a straight line being (practically) nil.
The Solutions 249
'Take the longest side of the triangle and call it AB; and on that
side of it on which the triangle lies draw the semicircle AFB. Also,
with centres A, B, and distances AB, BA, draw the arcs BDC, AEC,
intersecting at C.
Then it is evident that the vertex of the triangle cannot fall outside
the Figure ABDCE.
'Also, if it falls inside the semicircle, the triangle is obtuse-angled; if
outside it, acute-angled. (The chance of its falling on the semicircle is
practically nil.)
area of semicircle
'Hence required chance = --------
area of Fig. ABDCE
'Now let AB = 2a: then area of semicIrcle = It;'; and area of
Figure ABDCE = 2 x sector ABDC - triangle ABC;
= 2 ( 4 ~ ' ) _ J3a1 = a{ :n - J3}
n/2
... chance = ----
~ n - J3
3
3
8
6J3·
n
[Carroll, 1958, problem 58]
242. The clock that is losing time is correct once every two years,
whereas the stopped clock is right twice a day, every time that 'its'
time comes round!
[Carroll, 1961, p. 6]
243. Ten. Adding the four percentages together, the total is 310 per
cent. Distributing them as evenly as possible, three of each of the 100
250 Penguin Book of Curious and Interesting Puzzles
per cent total, there remains at least 10 per cent with all four disabili-
ties.
[CuthweIlis, 1978, p. 9)
244. ~ minutes.
[Cuthwellis, 1978, p. 9)
245. When the traveller crosses the International Date Line, which
was internationally agreed with just such a purpose In mind, but only
in 1884, long after the question had first troubled Carroll.
[Carroll, 1961, p. 4)
246. Provided friction is neglected, the weight at the other end of the
rope rises also, to match the monkey. Given friction in the pulley
wheel, the weight will move up less than the monkey, or indeed not
at all, if the pulley is sufficiently stiff.
[Carroll, 1961, p. 268)
247. The answer is always £12 18s lId, whatever the initial sum
chosen.
[Carroll, 1961, p. 269)
248. Assume, as is necessary but also implied, that when one basket
is within reach of the window, the other is at ground level. Raise one
basket, place the weight in it and lower the weight, raising the second
basket, into which the son steps and descends to the ground.
Lower the top basket, containing the weight. The son steps into
this basket at ground level and the daughter descends in the other
basket, raising the son and the weight.
Lower the weight again; the son descends against it, and the
daughter gets into the same basket as the son. The queen gets into the
basket containing the weight and descends against the weight of son
and daughter.
The daughter steps out at the top, before the queen steps out of the
bottom basket; the son remains and descends against the weight. The
son steps out and the weight descends to the ground. The son gets in,
and the daughter descends against son and weight.
Finally, the son gets out, the weight descends, and the son goes
down in the other basket, against the weight.
[Carroll, 1961, p. 318)
The Solutions 251
249. The amounts are equal. Moreover, it makes not the slightest
difference whether the water and brandy were stirred a little or a lot,
or whether the glasses contained equal quantities of liquid initially.
The quantity of liquid in each glass at the end is the same as at the
start, and therefore what one has lost, the other has gained.
[Hudson, 1954, Appendix A)
250. 'A level mile takes i hour, up hill}, down hill t. Hence to go and
return over the same mile, whether on the level or on the hill-side,
takes! an hour. Hence in 6 hours they went 12 miles out and 12 back.
If the 12 miles out had been nearly all level, they would have taken a
little over 3 hours; if nearly all up hill, a little under 4. Hence 3! hours
must be within! an hour of the time taken in reaching the peak; thus,
as they started at 3, they got there within t an hour of t past 6.'
[Carroll, 1958, p. 77)
251. For simplicity of calculation, turn the amounts given into modern
pence, so that the customer has 120, 24 and 6; the shopkeeper has 60,
12 and 1; and the friend has 48, 30, 4 and 3.
Then the customer gives 120 + 6 to the shopkeeper and 24 to the
friend; the shopkeeper gives 60 to the customer and 12 + 1 to his
friend; and the friend gives 30 + 4 to the shopkeeper, and 3 to the
customer. On balance the friend has gained or lost nothing, and the
shopkeeper is 87d better off, so the customer has given the shopkeeper
7s 3d.
[Carroll, 1961, p. 317)
[Cuthwellis, 1978, p. 14)
252 Penguin Book of Curious and Interesting Puzzles
253. 'The cat wins, of course. It has to make precisely 100 leaps to
complete the distance and return. The dog, on the contrary, is
compelled to go 102 feet and back. Its thirty-third leap takes it to the
99-foot mark and so another leap, carrying it two feet beyond the
mark, becomes necessary. In all, the dog must make 68 leaps to go
the distance. But it jumps only two-thirds as quickly as the cat, so
that while the cat is making 100 leaps the dog cannot quite make 67.
'But Barnum had an April Fool possibility up his sleeve. Suppose
that the cat is named Sir Thomas, and the dog is female! The phrase
"she makes three leaps to his two" would then mean that the dog
would leap 9 feet while the cat went 4. Thus when the dog finishes
the race in 68 leaps, the cat will have travelled only 90 feet and 8
inches.'
254.
[Loyd, 1959, Book 1, No. 14]
853
7491638897
5992
[Loyd, 1959, Book 1, No. 41]
3969
3745
2247
2247
255. 'In this remarkable problem we find that the lake contained
exactly 11 acres, therefore the approximate answer of "nearly 11
acres" is not sufficiently correct. This definite answer is worked out
by the Pythagorean law, which proves that on any right-angle triangle
the square o( the longest side is equal to the sum of the squares of the
other two sides.
E I - - - - - - - - - ' ~
L - - - ~ 7 ~ - - ~ - - - - ~ ~ - - - - ~ B
The Solutions 253
'In the illustration ABD represents our triangle, AD being 9 units
long and BD 17, because 9 x 9 equals 81, which added to 17 x 17
(289) equals the 370 acres of the largest field. AEC is a right-angle
triangle, and the square of 5 (25) added to the square of 7 (49) shows
that the square on AC equals 74. CBF also is a right-angle triangle.
The square of its sides, 4 and 10, prove the square estate on BC to
equal 116 acres. The area of our triangle ADB is clearly half of 9 x
17, which equals 76.5 acres. Since the areas of the oblong and two
triangles can plainly be seen to be 65.5, we deduct the same from 76.5
to prove that the lake contains exactly 11 acres.'
[Loyd, 1959, Book 1, No. 36]
256. 'The black pieces of paper are nothing but a delusion and a
snare. The pieces are placed to make a little white horse in the centre
as shown.
'It was this trick of the White Horse of Uffington which popularized
the slang expression, "Oh, but that is a horse of another colour!'"
[Loyd, 1959, Book 1, No. 45]
257.
[Loyd, 1954, Book 1, No. 102]
254 Penguin Book of Curious and Interesting Puzzles
258. Move the pieces in this sequence: 14 11 1287610 12874364
7 14 11 15 13 9 12 8 4 10 8 4 14 11 15 13 9 124 8 5 4 8 9 13 14 10 6 2
1. This is forty-four moves, the minimum possible.
The original puzzle is impossible to solve. All the possible positions
in which the fifteen titles and the single space might be arranged, can
be divided into two equal classes, call them the 'odd' and the 'even'
positions. From an 'odd' it is possible to reach any other 'odd'
position, but quite impossible to reach any 'even' position.
In particular, if in any particular position one pair of tiles is
swopped, the new position can never be obtained from the original
position. If, however, two such swops are made, the position is obtain-
able.
[Loyd, 1959, Book 1, No. 21]
259. The boy is five years old.
[Loyd, 1959, Book 2, No. 86]
260. Five odd 'figures' will add up to 14 as follows:
11
1
1
14
[Loyd, 1959, Book 2, No. 69]
261. If we let x be the bridge's length in feet, then the cow stands
!x - 5 from one end and !x + 5 from the other. The train is 2x from
the nearest end. The cow can travel (ix - 5) + (fx + 41) in the same
time that the train travels (2x - 1) + (3x - 1). These two periods of
time reduce to (x - 1) and 5(x - 1), so we see that the train is five
times faster than the cow. With this information we write the equa-
tion:
2x - 1 = 5(fx - 5)
This gives x, the length of the bridge, a value of 48 feet. The actual
speed of the train plays no part whatever in this calculation, but we
need to know it in order to learn the speed of the cow. Since we are
told that the train travelled at 90 miles per hour, we know the cow's
gait to be 18 miles per hour.
[Loyd, 1959, Book 2, No. 166]
The Solutions 255
262. The cheapest way to make an endless chain out of the six five-
link pieces is to open up all five links of one piece, then use them for
joining the remaining five pieces into an endless chain. The cost of
this would be $1.30, which is 20 cents cheaper than the cost of a new
endless chain.'
[Loyd, 1959, Book 2, No. 25)
263.
[Loyd, 1959, Book 1, No. 51)
264. 'Let 1 be the length of the army and the time it takes the army
to march its length. The army's speed will also be 1. Let x be the total
distance travelled by the courier and also his speed. On the courier's
forward trip, his speed relative to the moving army will be x - 1. On
the return trip his speed relative to the army will be x + 1. Each trip
is a distance of 1 (relative to the army), and the two trips are
completed in unit time, so we can write the following equation
1 1
--+--=
x-I x+l
This can be expressed as the quadratic: x
2
- 2x - 1 = 0, for which
x has the positive value of 1 + j2. We multiply this by 50 to get the
final answer of 120.7 miles.
'[In Part II) the courier's speed relative to the moving army is x-I
on his forward trip, x + 1 on hiS backward trip, and F-=I on
his two diagonal trips. (It does not matter where he starts his round
trip, so to simplify the problem we think of him as starting at a rear
corner of the square instead of at the centre of the rear.) As before,
each trip is a distance of 1 relative to the army, and since he
completes the four trips in unit time we can write:
256 Penguin Book of Curious and Interesting Puzzles
1 1 2
--+--+ =
x-I x+l Jx2=1
This can be expressed as the fourth degree equation: X4 - 4x' -
2Xl + 4x + 5 = 0, which has only one root that fits the problem's
conditions: 4.18112. This is multiplied by 50 to get the final answer of
209.056 miles.'
[Loyd, 1959, Book 2, No. 146)
265. Martin Gardner comments, 'Loyd's Cyclopaedia does not
answer this difficult problem ... The best procedure, supported by
the answers to similar problems in Dudeney's puzzle books, seems to
be the following:
'C, the slowest walker, always rides the tandem. He and A, the
fastest walker, ride the tandem for 31.04 miles while B is walking. A
dismounts, and C turns around and rides back to pick up B at a spot
5.63 miles from the start. Band C remain on the bicycle for the
remainder of the journey, arriving at the same time that A arrives on
foot. The total time is a little less than 2.3 hours.'
[Loyd, 1959, Book 2, No. 123)
266.
The Solutions 257
[Loyd, 1960, problem 144]
267.
[Loyd (jnr), 1928, p. 33]
268.
[Loyd (jnr), 1928, p. 17]
258 Penguin Book of Curious and Interesting Puzzles
269.
""'"
...... r--'
..;......t'"
........ i'
..............
[Loyd Onr), 1928, p. 19]
270. Call the two to-galion cans A and B. This is Loyd's solution, in
tabular form:
A B 5-qt 4-qt
10 10
5 10 5
5 10 4
9 10
9 10
4 10 5
4 10 2 4
8 10 2
8 6 2 4
10 6 2 2
[Loyd Onr), 1928, p. 21]
271. The landlady placed 20 cents on the counter to pay for H
pounds of bologna. Louis cut 21 pounds. She took Ii pounds for 15
cents and invested the remaining 5 cents in pickles.
[Loyd Onr), 1928, p. 25]
272. Seven pieces can be cut, though one of them is minute:
The Solutions 259
[Loyd (lnr), 1928, p. 26]
273. 'On Sunday, the first day of the week, Kate promised to marry
Danny "when the week after next is the week before last". Therefore
she will marry Danny in 28 days after her promise. Had she promised
a day earlier, then on Sunday, 22 days later, her promise would have
fallen due.'
[Loyd (lnr), 1928, p. 27]
274.
[Loyd (lnr), 1928, p. 29]
275. 'The girl weighed 11 H pounds when she arrived. She ate 1 ~
pounds of breakfast food and gathered 10 pounds of samples, which
increased her weight by 10 per cent.'
[Loyd (lnr), 1928, p. 33]
276. The historical incident was 'the dropping of the tea into the
sea', otherwise known as the Boston Tea Party.
[Loyd (lnr), 1928, p. 47]
260 Penguin Book of Curious and Interesting Puzzles
277. The Queen can be placed initially at any of the turning points in
the path.
i'..
,I
"-
t7 1/ '-.1'
i"-
~ ~ ~ 1/
~
"-
IL
~ ~ lX 12J
[7
X ~ ~
1/
"/
17
~ fSJ
~
" ~ 1/ /
1"'-
~
"-
17
,7:
"
~
17 I ~ I"- K
!2
"
#
I"-
[White, 1913, p. 42]
278. This is Loyd's solution, from Sam Loyd's Puzzle Magazine,
April 1908. A Queen's Tour from d1 (the white queen's initial square
in a game of chess) is impossible without resorting to non-chess
moves.
1\
\
~ !A
"-
\
I2V
I"
IZ J
"-
!L
1/
~ rs
LI
7
"t-.
1/ I ~
1'\
u l{
1"-
"g
The Solutions 261
279.
5
•
3 4
10
_-t.
6
_8 . ..ft.
7
v
.;,r
1h
12
14 1$
13 16 17
18
[White, 1913, p. 52]
280.
: ~
Ilz.)
~
til:
[White, 1913, p. 52]
281. (a) 1 c4 c5; 2 Qa4 Qa5; 3 Qc6 Qc3; 4 Q x B mate. Or 1 d4 d5;
2 Qd3 Qd6; 3 Qh3 Qh6; 4 Q x B mate.
(b) 1 e4; 2 Ke2; 3 Ke3; 4 Qf3; 5 Ne2; 6 b3; 7 Ba3; 8 Nd4+,
e x d mate.
(c) 1 f3 e5; 2 Kf2 h5; 3 Kg3 h4+; 4 Kg4 d5 mate.
(d) 1 e3 as; 2 Qh5 Ra6; 3 Q x as h5; 4 Q x c7 Rah6; 5 h4 f6;
6 Q x d7 + Kf7; 7 Q x b7 Qd6; 8 Q x b8 Qh2; 9 Q x c8
Kg6; 10 Qe6 stalemate.
262 Penguin Book of Curious and Interesting Puzzles
(e) 1 f4 e5; 2 Kf2 Qf6; 3 Kg3 and Black can force perpetual
check.
[White, 1913, pp. 58-9]
282. 'The illustration will show that the triangular piece of cloth may
be cut into four pieces that will fit together and form a perfect
square. Bisect AB in D and BC in E; produce the line AE to F making
EF equal to EB; bisect AF in G and describe the arc AHF; produce EB
to H, and EH is the length of the side of the required square; from E
with distance EH, describe the arc H1, and make 1K equal to BE;
now, from the points D and K drop perpendiculars on E1 to Land
M ...
'I exhibited this problem before the Royal Society, at Burlington
House, on 17th March 1905, and also at the Royal Institution in the
following month, in the more general form: "A New Problem on
Superposition: a demonstration that an equilateral triangle can be cut
into four pieces that may be reassembled to form a square, with some
examples of a general method for transforming all rectilinear figures
into squares by dissection."
'I add an illustration showing the puzzle in a rather curious
practical form, as it was made in polished mahogany with brass
hinges for use by certain audiences. It will be seen that the four pieces
form a sort of chain, and then when they are closed up in one
direction they form the triangle, and when closed in the other direction
they form the square.'
The Solutions 263
[Dudeney, 1907, No. 26]
283.
o t.
1
2
A
1Ht
•
FI.OOA
___ '_L_O_O_R __ __
A
t
•
...
3
1>
.... I

4)
f&'" "

I
.
e
,"FLOOR
[
...
.
•
FLOOR
&
'Imagine the room to be a cardboard box. Then the box may be
cut in various ways, so that the cardboard may be laid on the table. I
show four of these ways, and indicate in every case the relative
264 Penguin Book of Curious and Interesting Puzzles
positions of the spider and the fly, and the straightened course which
the spider must take without going off the cardboard. These are the
four most favourable cases, and it will be found that the shortest
route is in No.4, for it is only 40 feet in length. It will be seen that
the spider actually passes along five of the six sides of the room!'
[Dudeney, 1907, No. 75]
284. 'A very short examination of this puzzle game should convince
the reader that Hendrick can never catch the black hog, and that the
white hog can never be caught by Katrun.
'Each hog merely runs in and out of one of the nearest corners and
can never be captured. The fact is, curious as it must at first sight
appear, a Dutchman cannot catch a black hog and a Dutchwoman
can never capture a white one! But each can, without difficulty, catch
one of the other colour.
'So if the first player just determines that he will send Hendrick
after the white porker and Katrun after the black one, he will have no
difficulty whatever in securing both in a very few moves.'
[Dudeney, 1907, No. 78]
285. 'The diagram shows how the piece of bunting is to be cut into
two pieces. Lower the piece on the right one "tooth" and they will
form a perfect square, with the roses symmetrically placed.'
[Dudeney, 1907, No. 77]
286. The diagram shows how seven of the planks are used to 'round
off' the corner, so that the eighth plank can be used as a bridge to the
other side.
The Solutions 265
[Dudeney, 1907, No. 54]
287. This is the only possible arrangement. The casket is 20 inches
square.
12
'//..
~ ' -
[Dudeney, 1907, No. 40]
266 Penguin Book of Curious and Interesting Puzzles
288. This appears in 536 Puzzles and Curious Problems, though
another version, in which a dog runs back and forth between his
master and an approaching friend, had appeared in Modern Puzzles.
Question (1) is inserted merely in order to induce the reader to solve
question (2) by adding up a long series of fractions.
The fly first meets car B in 1 hour 48 minutes. Since the cars are
approaching each other at a combined speed of 150 miles an hour,
they meet after 3001150 = 2 hours, when the fly will be crushed.
(During that time the fly has been continuously flying in one
direction or the other, at 150 miles an hour. The fly has therefore
flown a total distance of 300 miles.)
[Dudeney, 1967, p. 26]
289. 'No doubt some of my readers will smile at the statement that a
man in a boat on smooth water can pull himself across with the tiller
rope! But it is a fact. If the jester had fastened the end of his rope to
the stern of the boat and then, while standing in the bows, had given
a series of violent jerks, the boat would have been propelled forward.
This has often been put to practical test, and it is said that a speed of
two or three miles an hour may be attained.'
[Dudeney, 1907, No. 52]
290. This is the simplest solution, by using an intermediate square.
-_..=_-=-:.
d
The Solutions 267
The crescent is in the form of two equal straight lines, a and b, joined
by two identical circular arcs. Making the cuts in the first figure, the
four pieces will form the square in the second, which is then dissected
by the dotted lines into the Greek Cross, with a total of ten pieces.
[Dudeney, 1907, No. 37]
291. This puzzle is typical of Dudeney's interest in properties of
numbers. 1,111,111 has only two factors, apart from itself and unity:
1,111,111 = 239 x 4649, both factors being prime.
So, either 239 cats caught 4649 mice each, or 4649 cats caught 239
mice each.
[Dudeney, 1907, No. 47]
292. 'The first player must place his first cigar on end in the exact
centre of the table. Now, whatever the second player may do through-
out, the first player must always repeat it in an exactly diametrically
opposite position.' In this way the first player can be certain of
always placing the last cigar.
[Dudeney, 1917, No. 398]
293. Dudeney placed the marks at 1,4,5, 14, 16,23,25 and 31 inches
from one end. He also gave another solution, with the marks at 1, 2,
3, 4, 10, 16, 22 and 28 inches from an end.
[Dudeney, 1926, No. 180]
294. 'The nine men, A, B, C, D, E, F, G, H, J, all go 40 miles
together on the 1 gall. in their engine tanks, when A transfers 1 gall.
to each of the other eight, and has 1 gall. left to return home. The
eight go another 40 miles, when B transfers 1 gall. to each of the
other seven and has 2 galls. to take him home.' This process is
repeated, until finally the last man, ], travels another 40 miles and
has 9 gallons to take him home, having travelled 360 miles out and
home.
[Dudeney, 1926, No. 49]
268 Penguin Book of Curious and Interesting Puzzles
295.
[Dudeney, 1917, No. 317]
296. With a choice of 125 or 100 yards, it is possible to go round in
just twenty-six strokes: the strokes required for all the holes are
evident, with the exception of hole 5, which is reached in three drives
of 125 yards and one stroke backwards of 100 yards.
[Dudeney, 1907, No. 32]
297. This is a variant of the Bridges of Konigsberg. There are just
two cells with an odd number of doors, and, in order to pass through
every door just once, it is necessary to start at one of these cells and
end at the other. Therefore, the route must start at the starred 'odd'
cell on the outside, and the fair demoiselle is at the other starred cell.
[Dudeney, 1907, No. 34]
The Solutions 269
298. 'Add together the ten weights and divide by 4 and we get
289 Ibs, as the weight of the five trusses together. If we call the five
trusses in the order of weight, A, B, C, D and E, the lightest being A
and the heaviest E, then the lightest 110 Ibs must be the weight of A
and B, and the next lightest, 112 Ibs, the weight of A and C. Then the
two heaviest, D and E, must weigh 1211bs, and C and E must weigh
1201bs. We thus know that A, B, D and E weigh together 2311bs,
which gives us the weight of C as 581bs. Now, by mere subtraction,
we find the weights of the other five trusses, 54 Ibs, 56 Ibs, 59 Ibs and
621bs, respectively.'
[Dudeney, 1917, No. 101)
299. This beautiful solution requires just five pieces and only two
cuts. The distance AB is equal to one half of the hypotenuse of the
triangle. The triangle should not be too large - Dudeney notes that if
it is larger than the square in area, then a dissection requires six
pieces.
[Dudeney, 1917, No. 152)
I
I
I
I
I
, I
,I
,I
-__ ,I
----i
270 Penguin Book of Curious and Interesting Puzzles
300. Only sixteen moves are required, and these are the only two
possible minimum solutions.
r-
---
--
--- -- ---
~ - -
-,
I
I
- --- -- --- -- -
I
I
r-
1- --
I
-- -}
I I
II
:
I I
-.1,
.!.
I
I
I I
I
:
I
I
I
I
I
:
I I
I
I
:
!
I
:
I
:, I
I
I I
I
I
L.., I I I I -.I
I I I
I
I
J .
I
L
I
!..
I
L -.I
--- --
_.
r ---
-- --- --
-- ---
,
I
_J
I
- --- -- ---
-- _.
--
I
I
r-
t---
--
---
--
I
-I
r-'
I
·1-
I
I
11
I
I ---
~ - -
-,
I I
I
I
L-!
:
I
--- -
I
I
I I
I
L
I I
1
---
--
_.
J
I
-L
---
~ - - ---
1- --
J
I
!...
---
1- --1-- r--I --- ---
J
[Dudeney, 1917, No. 320]
The Solutions 271
301. Three coins are placed as on the left, each touching the others,
and then two coins are added as in the second figure.
[Dudeney, 1917, No. 419]
302. De Morgan was born in 1806 and was 43 in the year 43' = 1849
(see problem 231).
Jenkins was born in 1860, was SZ + 6
2
= 61 in the year 54 + 6
4
=
1921, and 2 x 31 = 62 in the year 2 x 31' = 1922, and 3 x 5 = 15
in the year 3 x 54 = 1875.
[Dudeney, 1926, No. 23]
303. Nand N/(N - 1) have this property, whatever the integral
value of N, other than unity.
N ( 1) N
2
N+--=Nx 1+-- =--
N-l N-l N-l
[Dudeney, 1926, No. 93]
304. 'According to the conditions, in the strict sense in which one at
first understands them, there is no possible solution to this puzzle. In
such a dilemma one always has to look for some verbal quibble or
trick. If the owner of house A will allow the water company to run
their pipes for houses Band C through his property (and we are not
bound to assume that he would object), then the difficulty is got over,
as shown in our illustration.'
272 Penguin Book of Curious and Interesting Puzzles
[Dudeney, 1917, No. 251)
305. Arrange the six pennies as in the first figure. This can be done
exactly. Next move coin 6 as in the next figure. This is also exact.
Finally slide out 5 and place it against 2 and 3, and move 3 to just
touch 6 and 5.
This puzzle is usually presented as the problem of merely transform-
ing the first arrangement into the final arrangement, or sometimes of
transforming a triangle of six pennies (move 4 in the first figure to
touch 5 and 6) into the circle. Dudeney's presentation seems to me
much superior.
[Dudeney, 1926, No. 213)
The Solutions 273
306. Thirteen coins can be placed as shown in this figure:
[Dudeney, 1917, No. 429]
307. 'The secret of the bun puzzle lies in the fact that, with the
relative dimensions of the circles as given, the three diameters will
form a right-angled triangle, as shown by A, B, C. It follows that the
two smaller buns are exactly equal to the large bun. Therefore, if we
give David and Edgar the two halves marked D and E, they will have
their fair shares - one quarter of the confectionery each. Then if we
place the small bun, H, on the top of the remaining one and trace its
circumference in the manner shown, Fred's piece, F, will exactly
equal Harry's small bun, H, with the addition of the piece marked G
- half the rim of the other. Thus each boy gets an exactly equal share,
and there are only five pieces necessary.'
[Dudeney, 1917, No. 148]
274 Penguin Book of Curious and Interesting Puzzles
308. 'The reader will probably feel rewarded for care and pa-
tience that he may bestow on cutting out the cardboard chain. We
will suppose that he has a piece of cardboard measuring 8 in. by 2!
in., though the dimensions are of no importance. Yet if you want a
long chain you must, of course, take a long strip of cardboard. First
rule pencil lines B Band C C, half an inch from the edges, and also
the short perpendicular lines half an inch apart. Rule lines on the
other side in just the same way, and in order that they shall coincide
it is well to prick through the card with a needle the points where the
short lines end. Now take your penknife and split the card from A A
down to B B, and from D D up to C C. Then cut right through the
card along all the short perpendicular lines, and half through the card
along the short portions of B Band C C that are not dotted. Next
turn the card over and cut half through along the short lines on B B
and C C at the places that are immediately beneath the dotted lines
on the upper side. With a little careful separation of the parts with
the penknife, the cardboard may now be divided into two interlacing
ladder-like portions; and if you cut away all the shaded parts you will
get the chain, cut solidly out of the cardboard, without any join.
'It is an interesting variant of the puzzle to cut out two keys on a
ring - without join.'
1
________________________________ ____
Bf--· .. ··,--· .... r-- ..... r-- ..... - .... _ .... r-- ..... - .... &
__________________________________________
2
[Dudeney, 1917, No. 177]
The Solutions 275
309. 'The puzzle was to cut the two shoes (including the hoof
contained within the outlines) into four pieces, two pieces each, that
would fit together and form a perfect circle. It was also stipulated
that all four pieces should be different in shape. As a matter of fact, it
is a puzzle based on the principle contained in that curious Chinese
symbol the Monad.
'The above diagrams give the correct solution to the problem. It
will be noticed that 1 and 2 are cut into the required four pieces, all
different in shape, that fit together and form the perfect circle shown
in diagram 3. It will further be observed that the two pieces A and B
of one shoe and the two pieces C and D of the other form two exactly
similar halves of the circle - the Yin and Yang of the great Monad. It
will be seen that the shape of the horseshoe is more easily determined
from the circle than the dimensions of the circle from the horseshoe,
though the latter presents no difficulty when you know that the curve
of the long side of the shoe is part of the circumference of your circle.
The difference between Band D is instructive, and the idea is useful
in all such cases where it is a condition that the pieces must be
different in shape. In forming D we simply add on a symmetrical
piece, a curvilinear square, to the piece B. Therefore, in giving either
B or D a quarter >:.un before placing in the new position, a precisely
similar effect must be produced.'
[Dudeney, 1917, No. 160)
310. 'One object that I had in view when presenting this little puzzle
was to point out the uncertainty of the meaning conveyed by the
word "oval". Though originally derived from the Latin word ovum,
an egg, yet what we understand as the egg-shape (with one end
smaller than the other) is only one of many forms of the oval; while
some eggs are spherical in shape, and a sphere or circle is most
276 Penguin Book of Curious and Interesting Puzzles
certainly not an oval. If wt; speak of an ellipse - a conical ellipse - we
are on safer ground, but here we must be careful of error. I recollect a
Liverpool town councillor, many years ago, whose ignorance of the
poultry-yard led him to substitute the word "hen" for "fowl", remark-
ing, "We must remember, gentlemen, that although every cock is a
hen, every hen is not a cock!" Similarly, we must always note that
although every ellipse is an oval, every oval is not an ellipse. It is
correct to say that an oval is an oblong curvilinear figure, having two
unequal diameters, and bounded by a curved line returning into itself;
and this includes the ellipse, but all other figures which in any way
approach towards the form of an oval without necessarily having the
properties above described are included in the term "oval". Thus the
following solution that I give to our puzzle involves the pointed
"oval" known among architects as the "vesica piscis".
The Solutions 277
'The dotted lines in the table are given for greater clearness, the
cuts being made along the other lines. It will be seen that the eight
pieces form two stools of exactly the same size and shape with similar
hand-holes. These holes are a trifle longer than those in the school-
master's stools, but they are much narrower and of considerably
smaller area. Of course 5 and 6 can be cut out in one piece - also 7
and 8 - making only six pieces in all. But I wished to keep the same
number as in the original story.
'When I first gave the above puzzle in a London newspaper, in
competition, no correct solution was received, but an ingenious and
neatly executed attempt by a man lying in a London infirmary was
accompanied by the following note: "Having no compasses here, I
was compelled to improvise with the aid of a small penknife, a bit of
firewood from a bundle, a piece of tin from a toy engine, a tin tack,
and two portions of a hairpin, for points. They are a fairly serviceable
pair of compasses, and I shall keep them as a memento of your
puzzle" ,
[Dudeney, 1917, No. 157]
311. The sum is 9567 + 1085 = 10652.
[Strand Magazine, July 1924]
312. 'The first step is to find the distance travelled by the spiders.
We use the formula
DIStance = Velocity x Time.
According to our stated conditions, the velocity is 0.65 miles per
hour; but we want to get this in inches per second, since the time in
our problem is given in seconds and part of the distance in inches. Cal-
culating:
0.65 (mile)
1 (hour)
0.65 x 5280 x 12
60 x 60
gives us our velocity. Then
41,184 (inches)
3600 (seconds)
41,184 625
Distance = -- x - = 650 inches
3600 11
That is how far each spider travelled. But we were asked to find the
dimensions of the room. Therefore, now, let us spread out its walls,
ceiling, and floor onto a plane, much as if we were to open out the
six faces of a cardboard box to make one flat piece. Since there are
278 Penguin Book of Curious and Interesting Puzzles
F
\
\
\
\
\
E
\
'\
"
A
III
"
:Er-----,F
H
I \
G
I
\
I \
I \
,
I
,
, ,
I
,
" f
F'
o c
8 F
\
,
,
,
,
I
,
c
.
,
E ,
,
,
I
,
,
,
I
several possible paths, this must be done in all possible ways, keeping
the wall from which the spiders start fixed, and laying down the
others in such a way as to keep every face attached to another along a
common edge (see diagram). We can then see that the following eight
paths are possible:
The Solutions 279
1. Over the floor and parts of the two end walls.
2. Over the ceiling and parts of the two end walls.
3. Over one side wall and parts of the two end walls.
4. Over the other side wall and parts of the two end walls.
5. Over the ceiling and one side wall and parts of the two end walls.
6. Over the ceiling and the other side wall and parts of the two end
walls.
7. Over the floor and one side wall and parts of the two end walls.
8. Over the Roor and the other side wall and parts of the two end
walls.
Denote the three dimensions of the room by I = length, w = width,
and h = height. The lengths of the various paths may then be ex-
pressed:
Distance = I + h = 650 inches.
Distance = J 160' + (I + w)' = 650 inches
J(
h + w )' ( h + w )'
D,stance = -2- + 80 + 1 + -2- - 80
= 650 inches
(1,2)
(3,4)
(5,6,7,8)
To eliminate the radical signs, we may rewrite these equations:
(I + W = 650'
160' + (I + w)' = 650'
(h : w + 80 Y + (I + h : w - 80 Y = 650'
From these we find:
1 + h = 650
(I + w)' = 650' - 160' = 396,900 = 630'
l+w=630
Adding (I) and (II), we have
21 + h + w = 1280
whence
h + w
1 + -- = 640.
2
The third equation now yields
(1,2)
(3,4)
(5,6,7,8)
(I)
(II)
280 Penguin Book of Curious and Interesting Puzzles
e : w + 80)' = 650' - (640 - 80)' = 650' - 560'
108,900 = 330'
whence h + w = 500
From half the sum of equations (I), (II) and (III), we get
1+ h + w = 890
from which we obtain, by subtracting (III), (I) and (II) in turn,
I = 390, w = 240, h = 260'
[Kraitchik, 1955, pp. 18-19]
(III)
313. This is essentially the only solution. Note the rough symmetry
about the central column when the hexagon is in this orientation.
[Gardner, 1971, pp. 22--4]
314. The level fell. As long as it was in the boat, the cannon
displaced its own weight of water. After it sank, it displaced only its
own volume, which is a smaller quantity of water.
[After Williams and Savage, 1946]
The Solutions 281
315. The sum can change because the three numbers at the vertices,
and no others, are counted twice.
[Perelman, 1979, No. 49]
316. The front wheels are usually smaller in diameter than the back,
and therefore turn a larger number of times, and wear and tear on the
front axle is greater.
[Perelman, 1979, No. 65]
317. No it is not, because the single small cube at the centre of the
original large cube has six faces, and a separate slice is needed to
create each of those faces. Therefore six cuts is a minimum.
[Perelman, 1979, No. 122]
282 Penguin Book of Curious and Interesting Puzz/ps
318. They had counted the same. Tom, as he walks to and fro, will
meet some passers-by sooner, because he is approaching them, and
others with a delay, because he is walking away from them, but by
the time he returns to Mike, everyone who passed Mike will have
passed Tom also.
[Perelman, 1979. No.4]
319. 'We shall indicate the divisor by D, the quotient by Q, the digits
of the divisor by d" d ..... , d" and those of the quotient by
q" q .. ... , qlO' Hence, it is given that q. = 7, while it is clear at once
that qR = O. The digits ql and q7 must both be larger than q. (hence
larger than 7) and less than q. and q,; consequently, ql = q7 = 8 and
q. = q, = 9. From q7 = 8 it follows that 8 x D is less than
10,000,000 and at least equal to 10,000,000 - 97,999 = 9,902,001;
hence D must be less than 1,250,000 and greater than 1,237,750, so
that d, = 1, d. = 2, and d
1
= 3 or 4. From this it follows further
that qs = 8. Since the fourth digit of qs x D, thus of 8 x D, is a 7,
this shows that d. = 4 or 9, and that d, is at most 4. The assumption
that d
1
= 3 leads to d. = 9 (because D is greater than 1,237,750),
from which it follows (in connection with the third digit of q. x D
being 7) that we must have q. = 2 or 7. From the thirteenth row of
the division sum it is evident that (800 + q. + 1) x D is a ten-digit
number. However, 803 x d,d
7
is less than 803 x 1,240,000, hence
less than 995,720,000, so that q. = 2 drops out, and only q. = 7
remains to be examined. The second digit of the product obtained
when 7q'0 (that is, the number written with the digits 7 and q,0) is
multiplied by D is a 7; but we have
7q'0 x 1,239,7d,d
7
= 86,779,000 + (q,0 x 1,239,7d,d
7
) + (70 x d,d
7
)
the second digit of this number is not 7 for any of the possible values
1, 2, ... , 8 of q,0, so that q. = 7 is not possible, either, which makes
d
1
= 3 drop out. So we must have d
1
= 4, and D = 1,24d.,7d,d
7
(where d. = 4 or 9). The third digit of q. x 1,249,7d,d
7
is not a 7 for
any of the possible values of q., so we must have d. = 4. From the
fact that the third digit of q. x 1,244,7d,d
7
is a 7 it follows that q. =
4. The second digit of
4q'0 x 1,244,7d,d
7
= 49,788,000 + (q,0 x 1,244,7d,d
7
) + (40 x d,d
7
)
is a 7, from which it follows that q'0 = 6. The seventh digit of
898,046 x 1,244,7d,d
7
= 1,117,797,856,200 + 898,046 x d,d
7
is a 7,
from which we can deduce (since d,d
7
is less than 50) that d,d
7
=
The Solutions 283
k x 11, where k = 0, 1, 2, 3, or 4. Hence, for the dividend Q x D
we find:
q" 987,898,046 x 1,244,7d.d
7
= (q, x 1,244,700,000,000,000)
+ 1,229,636,697,856,200 + (k x 10,866,878,605)
+ (q, x k x 11,000,000,000)
The seventh digit of this number is the last digit of 6 + k X (q, + 1).
This has to be a 7, by the terms of the problem, so that k x (q, + 1)
must end in 1. In connection with k = 0, 1, 2, 3, or 4 (keeping in
mind that q, is different from 0), it follows that k = 3, hence d. =
d
7
= 3, and q, = 6. Consequently
D = 1,244,733 Q = 6,987,898,046
The dividend is then found as the product
Q x D = 8,698,067,298,491,718'
[Schuh, 1968,pp.319-20]
320. Arranged as in the diagram, the top brick overhangs the second
brick by one half of a brick length. The centre of gravity of the two
1
2
1
4
1
6
1
8
I
284 Penguin Book of Curious and Interesting Puzzles
bricks together is at G, so that they can together overhang the third
brick by i brick length. The first three can overhang the fourth by i
brick length and so on.
The total possible overhang is the sum of the series:
i+i+:+i+t.J+···
which increases without limit, being just one half of the harmonic
series I + t + i + i ...
With just four bricks, the maximum overhang is t + 1 + ! + 1 =
If.-, so that the top brick completely overhangs the bottom brick.
If the bricks are replaced by a pack of fifty-two cards, the maximum
overhang is a little more than 2 ~ card lengths.
321. With the method of the last puzzle, an overhang of at most If.- is
possible with four bricks. In the following arrangement the overhang
is (15 - 4j2)/8, or a little over 7/6, and thus a little more than in the
previous puzzle.
--------------+- C-+
+-d-+
___________ .....JI+- a -+ +- b -+
[Ainley, 1977, p. 8)
322. My birthday is on 31 December, and today is I January.
323. Neither the diameter of the cylinder or of the sphere are given.
If the problem is genuine, therefore, and not impossible to solve, it
must be because the diameter of the cylinder and the diameter of the
sphere are irrelevant. In other words, if a cylindrical hole of length 6
inches is to be cut out of a large sphere, then the two ends of the
hole must be close to an 'equator' of the sphere, and the hole very
wide, the small amount of material left will be equal in volume to
that left if a very thin and narrow cylinder of length 6 inches is cut
from a sphere which is only slightly over 6 inches in diameter. If this
is so, then you will get the same result if you cut a cylindrical hole of
no width at all from a 6-inch sphere. Suppose therefore that the
cylinder has zero diameter, so that the diameter of the sphere IS 6
inches, then the volume remaining when the zero-diameter cylinder
has been drilled out will be the original volume of the sphere, 11tY =
361t.
The Solutions 285
This is in fact the correct answer, and is independent of the
measurements that were omitted.
There is a matching 'surprise' in two dimensions. If the longest
line segment that can be drawn in the space between two concentric
circles, without crossing the inner circle, is 6 inches long, what is the
area of the annulus? The area is rc x J2 = 9rc, regardless of the actual
dimensions of the circles.
324. The suggestion that the sums named ought to add up to £30 is
nonsense. The diners have spent a total of £27, of which £25 was for
the dinner, and £2 for the waiter.
325. On submersion, the lead and the iron weights will be supported
by the water, according to the quantity of water they displace. Since
iron weighs less than lead, the iron equal in weight to the lead will
displace more water. The iron will therefore rise and the lead sink.
326. 'If the slice is a plane one, then since the top and bottom faces
of the original cube are parallel, the two lines ac and bd will be
parallel. However, to discover whether two lines are parallel in a
perspective drawing is not trivial, so a little more cunning is required.
a
\
\
\
\ f
\ /
\ I /
\i/
\ ~
c
286 Penguin Book of Curious and Interesting Puzzles
'Reconstruct the missing edge, E. Then ab will meet E, because
both lines lie in one face of the cube. Similarly, cd meets E. 'IF abcd is
a plane slice, then ab and cd will meet. In this case all three lines must
meet in the same point, because all three lines do not lie in the same
plane.
'Checking in the solution diagram, they do indeed appear to meet in
the same point and so the slice could be a plane one. If it is not, it is
only because its surface is not plane although its edges lie in one
plane.'
[Wells, 1979, problem 60]
327. One or other, or both, of the governments of Monia and Moria,
and therefore, indirectly, their tax-payers. Should you find yourself in
the fortunate position of this young man, you would indeed find that
'money grew on trees' as long as the contradictory exchange rates con-
tinued.
[After Northrop, 1960, pp. 8-9]
328. Only twenty-three people need be in the room, a surprisingly
small number. The probability that there will not be two matching
birthdays is then, ignoring leap years,
365 x 364 x 363 x ... x 343
36513
which is approximately 0.493. This is less than t, and therefore the
probability that a pair occurs is greater than 50--50.
With as few as fourteen people in the room the chances are better
than 50--50 that a pair will have birthdays on the same day or on
consecutive days.
329. Mary suggested that she take white in one game and black in
the other. She then played as white the moves that were played
against her by the master taking white, and for her replies in the same
game, she played the moves that the master playing black made to
her. The masters thus played each other, in effect, and in between
laughing managed to play a decisive game, giving Mary her 50 per
cent score.
[After Kraitchik, 1955]
330. The man must have shot the bear from the North Pole. The
bear was therefore a polar bear, and its colour was white.
The Solutions 287
Phillips's original problem was 'Polar conundrum: Starting from
the North Pole, I walk 40 miles due south, and then 30 miles due
west. How far am I now from the North Pole?' The deliberate choice
of 30 and 40 miles suggests a 3-4-5 right-angled triang!e and the
mistaken answer 50 miles.
331. Mrs Agabegyun might live at the North Pole. However, it is also
possible that she is living in Antarctica, rather near the South Pole.
Walking 5 miles south, she is very close to the pole, so near that her 5
mile walk in an easterly direction takes her a whole number of times
round the South Pole, bringing her back to the point she reached by
walking 5 miles south: the finalS miles north returns her to base.
In this case, there are an infinite number of solutions, because the 5
mile journey east may take her once round the South Pole, twice,
three times ... !
332. One red card and one blue one. It is not possible to say,
however, whether the red card or the blue was shown first.
[After PhiIIips, 1937, problem to]
333. The boy with the dirty face sa"w his companion with a clean
face, and suspected nothing. His companion, seeing the other with a
grimy face, assumed that his face was also dirty, and went to wash it.
[Phillips, 1932, 'Time Tests of Intelligence', problem 12]
334. He sees that the other Wise Men are both laughing. But if they
could see that his face was clean, they would realize at once on
looking at each other that they were victims themselves. Since they do
not, he must suppose that each is laughing at him. Therefore his
forehead is marked too.
335. Four rungs remain submerged, because the ladder rises with the
ship, which rises with the tide.
[PhiIIips, 1936, problem 0.1]
336. The engine-driver is Brown. Mr R lives at Leeds and either Mr
B or Mr 1 in London. One third of Mr ]'S income could not be the
guard's. So the guard's neighbour must be Mr B. So Mr 1 lives in
London and is therefore the guard's namesake, i.e., the guard is J.
Now B is not the fireman and so must be the engine-driver.
['Puzzle Pages Editorial', Games and Puzzles magazine,
Nos. 33-4, Feb-March 1975]
288 Penguin Book of CurIOus and Interesting Puzzles
337. 'Stritebatt's average of 30 is obtained by dividing his total of
runs scored by the number of times he was Out. Let him be Out m
times, and let him be Not Out n times. Then 12n runs more would
give him an average of 35, i.e. 5 more, and 12n = Sm. It follows that
the number of completed innings is 17 or some multiple thereof, i.e.
17, 34, 51, etc. But if Stritebatt has 34 innings, he scores on 32
occasions, and since his lowest score is 17 and no two scores are the
same, his lowest possible scores are:
17 18 19 20 ... 48
and since he is Not Out 10 times, his average must be at least 43.3. If
he has 51 innings, his average must be higher. Hence he has 17
innings only and his scores are: 0, 0, 17, 18, 19 ... 31. (For 360, the
total number of runs scored, can only be made up in this way.) His
best score, therefore, is 31 Not Out.'
[Phillips, 1934, problem 18]
338. '1. There are five lodgers; hence the number of rashers originally
on the dish must be 5, or 10, or 15, etc.
2. The number is 5. For the greatest possible original number would
be reached in the following way: Smith takes f rasher; Jones takes 1,
total t; Evans takes 3, total ~ ; and then either:
(a) Brown takes 1/, total ¥; Robinson takes i, total *t.
(b) Robinson takes i, total 6; Brown takes If, total ¥.
'Both these are less than 10; hence the original number of rashers is five.
3. Since Evans always leaves at least one rasher he cannot immediately
precede Smith; so there are three possibilities:
(a) Jones precedes Smith and takes 1 rasher; total i.
(b) Brown precedes Smith and takes i rashers; total 2.
'In both these Evans must take three rashers, leaving only f or ° for
the other two; this is impossible.
(c) Robinson precedes Smith and takes f rasher; total 1.
'Evans must precede Robinson; for if not he must take 3 rashers,
Jones must take 1, and there will be none left for Brown. Let Evans
take x rashers, then either:
(1) Jones takes 1 rasher, total x + 2; Brown takes ~ + t, total¥- +¥.
'For this to be 5 we must have x = i.
(2) Brown takes j + 1. total ¥ + 2; Jones takes 1, total ¥ + 3.
'For this to be 5 we must have x = t.
'Hence Evans takes one and half rashers of bacon.'
[Phillips, 1934, problem 20, contributed by j. W. Frame]
The Solutions 289
339. 'One of the four quoted statements is true.
'(I) Let the first statement be true. Then we have two hatters (Mr D
and Mr B). So this hypothesis is "out".
(2) Let the second statement be true. Then Mr G is d; Mr B is h; Mr
H is b. It follows that Mr D is g.
(3) Let the third statement be true. Then Mr H is b. So Mr D is
neither b, h nor d, and, once again, must be g.
(4) Let the fourth statement be true. Now Mr B is h, Mr G is not h,
d or g, and so is b. Whence, as before, Mr D is g.
'Hence while we cannot with certainty identify anyone of the
others, we know that Mr Draper is the grocer.'
[Phillips, 1950, 'Inference' problems, No. 15)
340. 19 represents 25. If the statement of the problem is simplified,
the answer is almost obvious: it states that ab x ab = cab.
[Phillips, 1960, 'One Hundred Elementary Problems',
problem 53)
341. (1) The raven's owner's feathered namesake must be a light-
coloured bird. Hence the raven is owned by one of the following: Mr
Dove, Mr Canary, Mr Gull, Mr Parrot. The first two of these are
bachelors and the raven is owned by Mr Gull's wife's sister's husband
- i.e. Mr Parrot owns the raven.
(2) Mr Crow owns a light-coloured bird, but Mr Crow's bird's
human namesake is married. Hence Mr Crow owns either the parrot
or the gull. But Mr Crow cannot own the parrot, for the parrot's
owner's feathered namesake is owned by the human namesake of Mr
Crow's bird; and Mr Parrot, we know, owns the raven. Therefore Mr
Crow owns the gull.
(3) Mr Raven must own the parrot, the gull or the dove. But Mr
Crow owns the gull, and if Mr Raven owns the parrot two people
would own the raven. Whence Mr Raven owns the dove and
(4) Mr Dove owns the canary.
(5) The crow's owner is unmarried; hence Mr Canary owns the
crow. Whence:
(6) Mr Starling owns the parrot and
(7) Mr Gull owns the starling.
[Phillips, 1960, 'Inference' problems, p. 165)
290 Penguin Book of Curious and Interesting Puzzles
342. A flat tax of 25s for each window, plus Is 6d for every square
foot of window.
[Hadfield, 1939, 'Puzzles and Problems by Caliban', problem 3)
343. This is Eddington's own solution, as printed in Phillips, 1960.
Denote the four male animals by A" Az, A
J
, A. and the females by
- A" - Azo - A
J
, - A •.
A child's mistake can be described as a transformation, e.g::
A" A
z
, A
J
, A. -+ - A" - Azo - AJ, - A.
The transformation is equivalent to multiplying by a matrix; thus
-1 0
o 1
o 0
o 0
We denote the matrices representing the misidentifications of the n
children by Em (m = 1, 2 ... n) so that the transformation is written
A-+Em A.
Owing to the condition in the second paragraph of the problem the
above transformation applies to the female as well as the male animals.
The condition in the third paragraph can now be written
Em{EpA) = - Ep{Em)A
[m being not equal to p)
so that
EpEm = -EmEp
The condition in the fourth paragraph is
Em
z
= 1 for boys Em Z = - 1 for girls.
(I)
(2)
It has been proved that no more than five fourfold matrices can
satisfy (I) and that (when the elements are real) three of them have
positive squares and two negative squares. More usually the theorem
is stated in the form that, with fourfold matrices, there cannot be
more than five mutually anti-commuting square roots of - 1, and
that three of them are imaginary and two real.
An actual set satisfying (1) and (2) is:
The Solutions 291
Boys
[ ~
1 0
n [ ~
0 0
J J [ ~
0 0
n
0 0 -1 0 0 -1
0 0 0 1 -1 0
0 1 0 0 0 0
Girls
[ - ~
1 0
JJ[ ~
0 0
~ J
0 0 0 -1
0 0 1 0
0 o -1 0 0
A non-diagonal element signifies a mistake of species. Thus the
only list with any names right is the second - a boy. He got two pairs
right, but interchanged the sexes of the other two pairs.
The answer is therefore:
(1) three nephews, two nieces
(2) a boy
(3) 4 completely right; 4 wrong sex.
(There are 'five other possible pentads besides the one given; but
they all have the same characteristic, that there is only one diagonal
matrix and it has two elements + 1 and two - 1. The sign of any of
the five matrices can be reversed. The answer is unaffected by these
variations.)
Appended is a specimen set of five lists which fulfil the conditions.
(T = Mr Tove, t = Mrs Tove, and so on)
Correct 1st 2nd 3rd 1st 2nd
Names Boy Boy Boy Girl Girl
T B T
J
B
J
B T b r
R
J
R b
i
B
J
R T R
t b t
i
b
i
b B R T R
r B
J
b
J
r T
344. It makes two complete rotations.
292 Penguin Book of Curious and Interesting Puzzles
345. No. Suppose that the chessboard is coloured, as is usual, with a
checkerboard pattern of alternating black and white squares. The
two squares removed from the opposite ends of a long diagonal will
be of the same colour. Suppose that they are both white. Then the
remaining squares are thirty-two black and only thirty white.
But when a domino is placed to cover a pair of adjacent squares, it
will inevitably cover one black square and one white square. Therefore
thirty-one dominoes can only cover thirty-one squares of each colour.
The removal of any pair of squares of the same colour makes a
covering impossible. However, if the squares removed are of different
colours, then a covering is always possible, as Gomory proved by this
'look-see' diagram:
The figure shows, as it were, a continuous rook's tour of the
board. Removing any pair of opposite coloured squares WIll splIt the
tour into two portions, each containing an even number of squares,
alternately black and white, whIch can be covered by dommoes.
[Black, 1952, p. 157)
346. On average half the women wIll bear a gIrl first, and half will
bear a boy, so on first bIrths, the numbers of boys and gIrls will be
equal. The women who bore a girl will then have no more children,
while the half who bore a boy will continue to bear, havmg as second
children, half boys and half girlS, so that the balance of boys and girls
is preserved.
Among the third children to be born within familIes, there will also
be a balance of boys and girls, and so on.
The Solutions 293
The fact is that the number of families which consist of only one
girl, which amounts to no less than one half of all families, will
exactly balance the much smaller number of families containing
several boys followed by a girl. Indeed, this amounts to no more than
the factthat! = 1 + i + to + ...
This solution assumes that the ratio of births of boys to girls is
indeed one to one; the ratio actually favours boys very slightly, but
this ratio in itself will never produce the surplus that the King
requires, and his ingenious scheme will be of no help at all.
[Gamow and Stern, 1958, p. 20, communicated by Victor
Ambartsumian, the eminent physicist]
347. '''Very simple, my dear Watson," the Sultan chuckled. "As a
matter of fact I expected this good news exactly on that day. My
people, as I suggested before, may be too lazy to organize the
shadowing of their wives for the purpose of establishing their faithful-
ness or unfaithfulness, but they have certainly shown themselves
mtelligent enough to resolve the case by purely logical analysis."
, "I do not understand you, Great Sultan," said the vizier.
, "Well, assume that there were not fony unfaithful wives, but only
one. In this case, everybody with the exception of her husband knew
the fact. Her husband, however, believing in the faithfulness of his
wife, and knowing no other case of unfaithfulness (about which he
would undoubtedly have heard) was under the impression that all
wives in the city, including his own, were faithful. If he read the
proclamation which stated that there are unfaithful wives in the city,
he would realize it could mean only his own wife. Thus he would kill
her the very first mght. Do you follow me?"
, "I do," said the vizier.
, "Now let us assume," continued the Sultan, "that there were two
deceived husbands; let us call them Abdula and Hadjibaba. Abdula
knew all the time that Hadjlbaba's wife was deceiving him, and
Hadjibaba knew the same about Abdula's wife. But each thought his
own wife was faithful.
, "On the day that the proclamation was published, Abdula said to
himself, 'Aha, tonight Hadjibaba will kill his wife.' On the other
hand, Hadjibaba thought the same about Abdula. However, the fact
that next morning both wives were still alive proved to both Abdula
and Hadjlbaba that they were wrong in believing in the faithfulness
of their wives. Thus during the second night two daggers would have
found their target, and two women would have been dead."
294 Penguin Book of Curious and Interesting Puzzles
• "I follow you so far," said the vizier, "but how about the case of
three or more unfaithful wives?"
• "Well, from now on we have what is called mathematical induc-
tion. I have just proved to you that, if there were only two unfaithful
wives in the city, the husbands would have killed them on the second
night, by force of purely logical deduction. Now suppose that there
were three wives, Abdula's, Hadjibaba's, and Faruk's, who were
unfaithful. Faruk knows, of course, that Abdula's and Hadjibaba's
wives are deceiving them, and so he expects that these two characters
will murder their wives on the second night. But they don't. Why? Of
course because his, Faruk's, wife is unfaithful, too! And so in goes the
dagger, or the three daggers, as a matter of fact."
• "0 Great Sultan," exclaimed the vizier, "you have certainly
opened my eyes on that problem. Of course, if there were four
unfaithful wives, each of the four wronged husbands would reduce
the case to that of three and not kill his wife until the fourth day.
And so on, and so on, up to forty wives."
• "I am glad," said the Sultan, "that you finally understand the
situation. It is nice to have a vizier whose intelligence is so much
inferior to that of the average citizen. But what if I tell you that the
reported number of unfaithful wives was actually forty-one?" ,
[Gamow and Stern, 1958, pp. 21-3]
348. The path of a point on the flange of a wheel moving on a rail
looks like this:
---------
--.............
Round about the bottom of each loop, the point is moving in the
opposite direction to its general motion. Therefore at anyone
moment, portions of the bottom of every wheel on the train are
moving, albeit temporarily, back to Bristol.
It is also possible, but less certain, that portions of a rapidly
moving wheel within the driving motor could be moving back to
Bristol, without even being flanged.
The Solutions 295
349. The bicycle moves backwards. However, the pedal moves for-
wards, relative to the bicycle, so that the pedals as a pair are rotating,
as would be expected, in the opposite direction to that required to
move the cycle forwards.
350. This is an example of the 'pigeon-hole' principle. Consider one
million boxes, numbered consecutively from 0, for the completely
bald, to 999,999 for those people (who might just exist according to
the information given) who have that many hairs on their head.
Place one slip for each person in the United Kingdom into the box
corresponding to his or her number of hairs. Then at least one box
must contain fifty slips of paper, corresponding to at least fifty people
with that same number of hairs on their head.
351. (a) Yes, you have, if a child skipping can be described as run-
ning.
[Wells, 1983-6, Series 1, problem 97]
(b) The maximum overlaps in each case have the same area, so three-
quarters of the triangle equals half the square, and the area of the
triangle is 24 square inches.
[Wells, 1979, problem 57, part 3]
(c) When passing on a spiral staircase, where the insides are narrower
and more difficult to walk on.
[Wells, 1979, problem 58, part 2]
(d) Tom thinks that Fred will not mind being five pence short, so
presumably Tom will not mind being five pence short himself, which
is what he would be if he gave Fred ten pence - which will naturally
also satisfy Fred.
[Wells, 1979, problem 57, part I]
352. The shortest route is shown overleaf. Most attempts to get
through the maze end up with Theseus unable to leave without
missing his last chance to turn left or right. His wander round the
lower right corner is necessary to change the parity with which he
approaches the north exit.
296 Penguin Book of Curious and Interesting Puzzles
[Wells, 1979, problem 54]
353. I had dashed on to the platform at the rear end of the first train;
I knew that the exit at my destination station was at the front end of
the train, so I walked down the platform and reached the other end
just as the next train arrived, and I got into the front carriage. Result:
I had quite simply walked the length of the train at the first station,
instead of at my destination station, and no time was wasted at all.
[Wells, 1979, problem 50]
354. Four men and four women were present. If you mark points for
individual men and women, joining dancing partners by lines, then
you will construct a skeleton of points and lines in which every region
has four edges and three edges meet at every point. All skeletons with
these properties are equivalent to either a single cube or to several
entirely separate cubes. Since the soiree is described as intimate, it is
reasonable to suppose that the skeleton is one cube only. There is one
person for each of its eight vertices.
[Wells, 1979, problem 42]
The Solutions 297
355. This is the shape her husband produced. The size and propor-
tions of the base are irrelevant as long as the base and sides of the
box are rectangles. It will be possible to choose the height of the cut
at each corner, and then make the cut with just one stroke of the
sword, so that by using the appropriate corner, either one, two, three
or four measures can be poured out. The technique is to fill the box
so that the rice has a level surface, just comes up to the lip of the
chosen corner, and leaves the opposite corner of the base rectangle
just showmg.
2h
rWells, 1979, problem 32]
356. Two pieces, cut as shown. These two squares can be interlaced
to form the mono
[Wells, 1979, problem 16]
298 Penguin Book of Curious and Interesting Puzzles
357. This is an ordinary overhand knot, composed of just twenty-
four cubes.
358.
[Doug Engel, in Games and Puzzles, No. 36, May 1975]
"
, '.
,,' ............
" .. )
,
,
,
,
,
, .
'.'
,
,
,
,
,
,
,
,
,
The Solutions 299
359. Three, one of the squares being the hollow square in the middle.
[After Adams, 1939, problem C12]
360. He lost. Winning multiplies his stake by 11; losing halves it. So
three wins and three losses In any order multiply his original stake by
27
64
So he loses * of his £1, or 58 pence to the nearest penny.
[After Adams, 1939, problem B5]
361. Ninety-nine blocks, each block requiring just ten chops.
[Adams, 1939, problem C71]
362. The Red Lion.
[Adams, 1939, problem B32]
363. It can be done in fourteen moves: (1) C-a; (2) B-d; (3) E-b-B-c;
(4) C-E-b-B; (5) A-c-C-a; (6) E-A; (7) C-c; (8) B-B-b-E; (9) C-B-d;
(10) E--c-B-b; (11) C-B-c-A-c; (12) E-B-c; (13) D-b-B-d; (14) B-b.
[After Adams, 1939, problem C170]
364. Whatever the shape of the original triangle, the central triangle
is ~ of it, in area.
The simplest solution is based on the idea that it is possible to
shear the original figure so that it becomes equilateral, without
changing the relative sizes of its constituent triangles.
'Draw dotted lines as in the figure. We use the proposition that
triangles with the same altitude are in proportion to their bases. Call
each of the three smallest triangles the unit of area. Then each of the
300 hnguin Book of Curious and Interesting Puzzles
triangles marked 3 has three times the unit area. Of the four remaining
triangles, mark the central one y and the others x. Then by the above
proposition, x + y + 3 = 3 (x + 1) and 2x + y + 7 = 3 (x + 5),
when x = 8, y = 16. Since the whole triangle = 52, the central
one = 4/13 of the original. (By comparing triangles, we also find that
each line from a vertex to the division point of the opposite side, IS
divided in the ratio 4: 8: 1.) By substituting n for 4 in this solution, we
find similarly that the central triangle = (n - 2)'/(n
'
- n + 1) of
the original (and each internal line from a vertex is divided into the
ratio n: n(n - 2): 1).'
[Graham, 1963, problem 52]
365. 69
'
= 4761 and 69
3
= 328509.
366. The dispute can be settled in the following manner: We give the
priority of choosing the piece of ham to the third co-owner. She will
choose, of course, the piece which according to her home balance is
not less than either of the remaining two pieces. That is the piece
whose value, according to her opinion, is not less than $4.00. Such a
piece must exist because, by division of the whole into 3 parts, one of
the parts cannot be less than t of the total weight.
'Afterwards the second woman chooses her piece. She must also be
satisfied because, after the third woman took her share, there re-
mained at least one piece which, according to the balance in the shop
on the corner, corresponded to a value not less than t.
'The first woman, who receives the remaining piece, must be
satisfied, since she considered all the pieces to be of equal weight.'
[Steinhaus, 1963, problem 49]
The Solutions 301
367. The same. It is perfectly possible for the shortest giant and the
tallest midget to be one and the same person. Indeed, if the men
formed an array of k columns and m rows, then any man who has at
least k - 1 colleagues shorter than himself and m - 1 taller than
himself, can be in that double position, for a suitable arrangement of
the men.
[After Steinhaus, 1963, problem 59)
368. Five. Suppose that it were possible for one town to be connected
to six other towns. It is connected to one other town which is the
closest town to itself. The other five connections are all to towns for
which it IS the closest other town. But if six points are arranged
round a central point, the central point can.only be equal closest to
the other points, when they are arranged at the centre and vertices of
a regular hexagon. As soon as the distances are adjusted, even very
slightly, to make them all different, one of the ring of points will be
nearer to another point than it is to the centre of the hexagon,
contradicting the information given.
[After Steinhaus, 1963, problem 71)
369. Five is sufficient for a gallery of twenty walls at right-angles.
The diagram shows a worst-case plan, which is nevertheless a
common gallery design. One guard is needed for each of the rooms
off the main corridor, and these guards can be placed so that the
walls of the corridor also are always surveyed.
370. Very easily! Here are some especially composed sample figures
to illustrate what can happen, adapted from Chapter 4 of Hugh
ApSimon's Mathematical Byeways in Aylmg, Beeling and Ceiling
where there is a full discussion of the 'paradox'.
Paul Frank
First half of year Runs 252 84
Times out 4
Average 63 84
302 Penguin Book of Curious and Interesting Puzzles
Second half of year Runs 84 252
Times out 3 7
Average 28 36
Whole year Runs 336 336
Times out 7 8
Average 48 42
Note the uneven distribution of Frank's scores and the fact that if his
scores for the first and second halves of the year had been switched,
then Paul would have led in the first half of the year, Frank in the
second, and the result would not seem at all paradoxical.
[ApSimon, 1984, p. 23)
371. It will definitely end up in one of the other three pockets. It is
not possible for it to bounce for ever or end up in the pocket it
started from.
Suppose that it starts from pocket A, that the sides are integers p
and q, as marked, and that 1 unit of distance is the length of the
diagonal of one square.
Then, in travelling back and forth between the left-hand vertical
edge and the right-hand vertical edge, the ball travels q, 2q, 3q ...
units, while in travelling back and forth between the top and bottom
edges it travels p, 2p, 3p ... units.
Therefore when a multiple of p first equals a multiple of q the ball
will have travelled a whole number of times between the vertical
edges and a whole number of times between the top and bottom
The Solutions 303
edges - which is another way of saying that it will be in one of the
corners.
Assuming that it stays in the corner, rather than bouncing back
out, it can never retrace its path and end up in the corner from which
it started.
[Mauldin, 1981, problem 147]
372. It must be struck parallel to one of the diagonals, and its total
path back to its starting point is double the length of a diagonal.
373. All the statements contradict one another, and therefore at most
one of them can be true, in which case the other nine statements will
be false, which is what statement nine asserts. Therefore statement
nine is the only true statement.
374. Standing on a bookshelf In the normal order, which we can
reasonably assume, volume one is on the left, its front flat against
the back cover of volume two, and the back cover of the third volume
is flat against the front cover of the second volume.
Therefore the bookworm only actually bores through the complete
second volume, a distance of 8 cm.
375. If a-be means move coin a to touch coins band e, then five
moves are reqUired from H to 0: 1-56, 3-14, 4-58, 5-23, 2-54. But
no less than seven moves are required to get back from 0 to H:
D-CE,G-CD,D-CG,G-BD,C-AG,A-BE,E-FH.
[Brooke, 1963, No. 40]
304 Penguin Book of Curious and Interesting Puzzles
376. Only three coins need be moved.
,
I \
I I
I I
" /
- - ~
377. Extraordinary to relate, it is done very easily. The numbers 1 to
9 as arranged in the usual magic square will do the trick.
A 6 1 8
B 7 5 3
c 2 9 4
If Alan's dice have the numbers in the first row, with opposite faces
of the dice showing the same number, and if Barry takes the numbers
in row B, and Chris those in row C, then A beats C, beats B, beats A,
each with the odds of 5 to 4. They could just as well take the columns
instead.
[Berlekamp, Conway and Guy, 1982, p. 778)
378. The distance is equal to OX. This IS obvious when the mirrors
are at 45° and the incoming ray is instantly reflected at right-angles to
the lower mirror, but the same result is true whatever the angle
between the mirror, provided that that angle is an integral fraction of
45°. If it is not, then the ray cannot emerge along its initial path.
[Mackie, A. G. and Jellis, G. P., in The Games and Puzzles
Journal, No.4, March-April 1988, p. 58)
The Solutions 305
379. The balloon swings left with the car. This is most easily under-
stood by considering that it is the air within the car that is heavy, and
the coal-gas in the balloon relatively light, and naturally it is the
heavier air that swings outwards, forcing the balloon inwards.
[Morris, 1970, p. 203, communicated by Gerald Stonehill]
380. The 20-litre barrel contains beer. The total quantity of wine is
divisible by 3, since it was bought in two parts, one double the other.
The total of the tens digits in the quantities is a multiple of 3, but the
sum of the unit digits is 29. Therefore, the barrel containing the beer
contains a multiple of 3, with 2 remainder. The 20 litre barrel is the
only possibility.
The first customer bought the IS-litre and 18-litre barrels, and the
second took the 16-litre, 19-1itre and 31-litre barrels.
381. 'Twice four and twenty' could be either 28 or 48, of which only
the first number is divisible by 7. Therefore Jill shot seven birds, and
it was these seven who remained, as the others flew away.
[Morris, 1972, problem 39]
382. The buses do indeed run very regularly, and at equal intervals,
every bus in one direction arriving, say, 1 minute after the previous
bus in the opposite direction, and many minutes before the next bus
in that same direction.
383. 16/64, 26/65, 19/95 and 49/98. There are many other such
cancellations with larger numbers, such as 143185117018560 = 1435/
170560 and 4251935345/91819355185 = 425345/9185185.
[Domoryad, 1963, p. 35]
384. You need turn over only two cards. The first card shows a
vowel, and so you must test whether it has an even number on the
reverse. The third card shows an odd number, and this would
contradict my claim if the letter on the other side were a vowel, so
you must test card three also. Neither of the other cards can affect my
claim either way.
385. He makes eight cigarettes and smokes them, leaving eight ends
from which he makes two more cigarettes, a total of ten.
386. 5 pence and 10 pence. One of them is not a 10 pence piece, but
the other is!
306 Penguin Book of Curious and Interesting Puzzles
387. If the level of beer in the can is above the centre of gravity, the
centre of gravity can be lowered by drinking more beer. Likewise, if
the level of the beer is below the centre of gravity then the centre of
gravity could be lowered by replacing some of the beer, to increase
the amount of beer below the centre of gravity line.
Therefore, the centre of gravity will be a minimum when neither of
the above cases applies, that is, when the surface of the beer and the
centre of gravity coincide.
The exact position of this can only be calculated when more data
are given about the size and weight of the can, and the density of the
beer.
388. Two, one on either side.
389. After twenty-nine days.
390. Every time a pair of delegates shake hands, the total number of
handshakes made increases by two. In other words, the total is
always an even number. If an odd number of delegates shook hands
an odd number of times, then however many shook hands an even
number of times, the grand total would be odd, which is impossible.
Therefore the conclusion of the puzzle follows.
391. No. The second man has not been placed anywhere.
392. 'Let A = J2.fi.. If A is rational, it is the desired example. On
the other hand, if A is irrational, then A.fi. = 2 is the desired exam-
ple.'
[Litton's Problematical Recreations, 1967, No.9, problem 4]
393. It is only necessary for the lines to cross on the shoe. This is a
simple and symmetrical solution.
The Solutions 307
394. The woman was buying individual house numbers from an iron-
mongers.
395. 'Since motion is relative, consider the hoop as fixed and the
poor girl whirling around. The original point of contact on the girl
traverses the diameter of the hoop twice, and this is the required dis-
tance.'
B ~ ~ - - - - - - - - ~ - - - - - - - - - - ~ D
In the diagram, C is the original point of contact. As the inner
circle rolls against the outer circle, C moves along the diameter BD.
[Trigg, 1985, problem 62]
396. John is a young lad and too short to reach the lift button for
any storey higher than the sixth.
397. Call the slices A, Band C. Fry A and B on one side each, and
then swop B for C and fry A on the other side and C on its first side.
Now swap A for B and fry the second sides of Band C at the same
time. Total time, 60 seconds.
398. The number of moves required is always n - 1, and cannot be
shortened by, for example, fitting together pieces in chunks, and then
joining the chunks together - though of course there are other good
reasons for tackling a jigsaw puzzle in that manner.
[Trigg, 1985, problem Q29]
308 Penguin Book of Curious and Interesting Puzzles
399. By similar triangles I = ~ , so xy = 1
Also (x + 1)2 + (y + 1)2 = 42 = 16
So (x + y)2 + 2(x + y) = 16
and x + y = - 1 ± JV
Since x + y is positive, x + y = JV -
and x - y = J(x + y)2 - 4xy = Jr-
14
---
2
-
JV
-;:I=7
SOX = HJ14 - 2JV + JV - 1) = 2.76
and the height of the top of the ladder above the floor (x + 1) IS
approximately 3.76 m.
i
x
11------\
+-y-+
400. The following solution was contributed by a reader to L. A.
Graham's The Surprise Attack in Mathematical Problems. It avoids
the difficulties to which this deceptively simple-looking puzzle (like
problem 399) so easily leads.
'One of our readers rose to this task as follows: "From the similar
triangles z = 8xlc and y = 8xld, where c and d are the left and right
segments of the required length x. Adding, we get z + y = 8x
'
lcd
and multiplying, zy = 64x
'
lcd or 8(z + y).
The Solutions 309
y
L - ~ - - - - - - - - - - - - - - - - - - - - ~ A
-./560
From the diagram, 900 - z' = 400 - y', from which z' - y' =
500. We represent this last identity by a triangle, as in this figure,
from which it is readily seen that cos A + cot A = 500(z + y)1 zy,
which from the previously derived relation equals %0/8 or 2.795l.
We now merely look up a trig table, or use a calculator, to find the
angle whose cosine and cotangent add up to the figure, and get A =
27° 38' 30" approximately, from which z = %o/cos A = 25.24 and
the required value of x = J900 - (25.24)' = 16.2 feet." ,
[Graham, 1968, problem 6]
401. The total number of edges meeting at all the vertices, if summed
vertex by vertex, is double the number of edges, and therefore an
even number. If an odd number of vertices had an odd number of
edges meeting at them, then the grand total would also be odd - a
contradiction. Therefore the answer to the question is 'no'.
402. If the street contains more than 100 houses but less than 1000,
then Mr Jones lives at No. 204, and the street is numbered from 1 to
288.
[After Beiler, 1964, p. 297, problem 23]
403. Thirty-six matches, because each match played eliminates one
player, and thirty-six players must be removed to leave one winner.
There are many ways to arrange the pairings, but the total number of
matches played is not affected.
404. He kept ducks.
405. A man was sitting on a three-legged stool eating a leg of ham,
when along came a dog and snatched the ham away. The man threw
the stool at the dog, and recovered his ham.
310 Penguin Book of Curious and Interesting Puzzles
406. The bottle costs 20tp and the cork fp.
407. Take a single sweet from the jar wrongly labelled MIXED. You
know that this jar is not mixed and so whatever sweet it contains tells
you its correct description. Suppose that it contains aniseed balls.
Then the jar wrongly labelled ANISEED BALLS must contain chocolate
drops and the other jar contains the mixture.
Finally, empty one of the jars, fill two of the jars in succession with
their correct contents, and replace the contents of the third jar, which
is almost certainly much easier than switching the labels round.
408. There is no knot. Try it and see.
[Budworth, 1983, p. 144)
409. 'We can build concentric hexagons containing 1, 6, 12, 18, 24,_
30, 36, and 42 circles.
'When Rlr (the ratio of the radius of the table to the radius of each
circle) becomes sufficiently large there will be room for extra circles
as indicated by" above. If there is an even number of Circles per side
in the last hexagon, an 'outsider' can be placed centrally if
The Solutions 311
1 + J3
2
R l r ~ -_-J3---::'3 i.e .• f R l r ~ 13·9
2
Two more "outsiders" can be put each side of this one if
[(R + r)z( ~ ' + (2r)'T + r:::;; R
R' R
i.e. if 0 :::;; - - 14 - - 15
r r
i.e. if 0 :::;; (; + 1) (; - 15)
i.e. if Rlr ~ 15
Hence in the example given three "outsiders" can be accommodated.
'The number of saucers that can be placed on the table is:
1 + 6 + 12 + 18 + 24 + 30 + 36 + 42 + (3 x 6) = 187'
[Kendall and Thomas, 1962, problem e12]
410. Two of the balls should be placed along one diagonal of the
lower part of the cube, and the other two along the other diagonal in
312 Penguin Book of Curious and Interesting Puzzles
the upper portion, so that the balls are at the vertices of a regular tetra-
hedron.
Looking down on the cube at the balls in the lower layer, the
length of the diagonal of the square is 2(5 + s.J'i), so the length of
one side of the cube is .J'i(S + s.J'i) or 10 + s.J'i, approximately
17.07 inches.
By symmetry, the remaining two balls will fit as described in the
top layer, just touching these two balls.
411. This beautiful argument is due to Guy David and Carlos Tomei.
Here is a full box. The calissons pointing in the three directions
have been shaded grey or black, or left white.
The shaded figure can be seen as a picture of a three-dimensional
arrangement of cubes, in which the black calissons are the top faces,
the white calissons are facing forwards, and the grey calissons face to
the right. The total number of calissons of each shade is just equal to
the total area of each face of the cube, and this figure is of course the
same in all three directions.
[American Mathematical Monthly, Vol. 96, No.5, May 1989]
The Solutions 313
412. He wraps the gun in canvas to make a rectangular package 2
metres long by at least 1.14 metres wide, so that the gun lies
diagonally across the package. It is now an acceptable packet.
[Kraitchik, 1955, p. 40, problem 50]
413. Sixteen extra cigarettes can be accommodated:
'Twenty cigarettes are placed in the bottom layer. In the second layer
instead of having twenty, we place nineteen, arranged as shown in the
diagram. Then we continue with alternate layers of twenty and nine-
teen.
Original method New method
'Suppose the diameter of a cigarette is 2 units. The second and
subsequent layers, using our new method, will add only 1.732 units to
the height. The depth of the box is 16 units, since it originally
contained eight layers. With our new method we shall get nine layers:
2 + 8 x 1.732 is equal to 15.856. So with five layers of twenty and
four layers of nineteen we shall get 176 cigarettes into the box.'
[Brandreth, 1984, p. 55]
414. Move A a little way away, and press down on coin B to stop it
moving. Flick A against Band C will move away, allowing A to be
placed between Band C.
415. 601 is less than 2
10
but more than 2
9
, so they played ten games.
The gross total of stakes was 1 + 2 + 4 + 8 + ... + 2
9
= 1023
cents. If the gross winnings were Steve x cents and Mike y cents, then
x + y = 1023 and x - y = 601; solving, x = 812 and y = 211.
Only the stakes in the first, second, fifth, seventh, and eighth games
add up to 211 cents. So those are the games that Mike won.
[Madachy, 1966, problem 7, p. 177]
314 Penguin Book of Curious and Interesting Puzzles
416. 'Solving the inequalities simultaneously, we find that 561
> N > 53:\. Knowing that N = 54, 55 or 56, the other inequalities
lead to the unique solution: P = 26, E = 19, N = 55.'
[Litton's Problematical Recreations, 1967, No. 11, problem 10)
417. From the numbers given it is clear that the number chosen
cannot start with a O. The number of allowable pandigital numbers is
therefore the total number, including initial zero, less the number
starting with a zero: it is 10! - 9! = 9 x 9!
There are 9,000,000,000 numbers in the range, so the probability is
(9 x 9!)/(9 x 10') = 362,88011,000,000,000 = 0.00036288, or less than
1 in 2500.
418. One quarter. This is proved by an elegant geometrical argument.
Consider an equilateral triangle, whose altitude is equal to the
length of the stick. For any point inside the triangle, the sum of the
perpendiculars from the point to the sides of the triangle is a constant,
equal to the altitude of the triangle, tAat is, to the length of the stick.
(This is also true for points outside the triangle, provided the appropri-
ate perpendiculars are counted as negative in length.)
If the point lies inside any of the shaded triangles, then one
perpendicular will be longer than the sum of the other two, and no
triangle is possible. Therefore, the divisions of the stick for which a
triangle can be formed correspond to the points in the central triangle,
which is one quarter of the whole.
419. One quarter of the length. The breaking point of the stick, since
the break is made at random, is equally likely to occur anywhere
along the stick, and so the shorter of the two pieces is equally likely
The Solutions 315
to be any length from zero to one half, and will on average be a quar-
ter.
(The stick can hardly be said to be broken if the break is at one
end, but the probability that it will be broken exactly at an end is
zero, and equal to the probability that it will be broken exactly at the
half way point.)
[Mosteller, 1987, problem 42]
420. There are five equal gaps between six strikes, so the time
between strikes is 3 seconds. When striking 12 there will be 3 x 11 =
33 seconds between the first and last strikes.
421. Let the distances of the four towns along the road, from some
fixed point, be a, b, c and d, in the order in which they are named.
Then the first instruction takes the treasure seeker to t(a + b) and the
second to Ha + b) + He - !(a + b)) = l(a + b + c).
By a similar calculation, the fourth instruction takes the seeker to
Ha + b + c + d), and this symmetrical expression can be calculated
without knowing which town is which.
In fact, it is the centre of gravity of the four points, and from the
distances given, is located at X, which is 1 mile from Band 7 miles
from C.
[NCTM, 1965, problem 91]
422. The grandfather is 66 and his grandson is 6.
[NCTM, 1965, problem 120]
423. The grandson was born in the twentieth century, in 1916, anc
was 16 years old in 1932. The grandfather was born in the n i n e t e e n t ~
century, in 1866, and was 66 in 1932.
[Perelman, 1979, problem 5]
424. A third diagonal can be added, to complete an equilatera
triangle, anyone of whose angles is 60°.
425. The circles are equally spaced and their areas are 1,4, 9, 16,25
36 and 49. Since 9 = 25 - 16, and 25 - 1 = 49 - 25, the inside of
circle C is equal In area to the annulus between circles D and E, and
the region between circles A and E equals in area the annulus
between E and G.
316 Penguin Book of Curious and Interesting Puzzles
426. 'Fred's bets were in the proportion 11(3 + 1), 11(4 + 1), 11(7 +
1), 11(9 + 1) and 11(39.+ 1), the last being placed on each of five
horses. These fractions add up to 0.8, and no matter which horse
wins, the winnings plus the returned winning stake total 1.00, i.e. the
profit is 0.2. Hence Fred's total stake was £800 and his profit £200.
'It's no use trying it yourself - real bookies fix the odds better than
that!'
Fred's stakes were: £250 on Bonnie Lass, £200 on Golden Stirrup,
£125 on Two's a Crowd, £100 on Greek Hero, and £25 on each of
the five others.
[Eastaway, 1982, p. 94]
427. Place the set square so the vertex at the right-angle lies on the
circle.
Because of the property that 'the angle in a semi-circle is a rIght-
angle', the line XY will be a diameter of the circle. By repeating the
process a second diameter is obtained, and the centre is their inter-
section.
428. 'Let A be the angle in minutes between the hands, with the
minute hand ahead of the hour hand. If the hour hand moves through
two-thirds of angle A, the minute hand must move through (60 - A) +
:tAo Since the minute hand moves twelve times as fast as the hour
hand, 12(M) = 60 - M and A = 6H minutes. The elapsed time is
12 x ~ x 6H = 55A minutes.
'It is very interesting that if the hour hand only moves through one-
third of the angle so that the minute hand ends up ahead of the hour
hand, an identical result is obtamed. While the hour hand moves
through one-third of the angle B, and the minute hand moves through
(60 - B) + 1B. Proceeding as before, 1 2 ( ~ B ) = 60 - iB and B = 13M
The Solutions 317
minutes. The elapsed time is now 12 x t x 13H- = 55f,- minutes.
'In order to rule the lines as soon as possible after 3 o'clock, the
angle chosen must be the smaller one, 6+1 minutes. Then the hour
hand movement X must be one-twelfth the movement of the minute
hand: 15 + 6+1 x X; 12X = 21+1 + X, and X = I l ~ ~ minutes. The
milllmum time to reach thiS configuration is 12 x I l ! ~ = 2 3 l ! ~ min-
utes.'
[Contributed by Marlow Sholander to Graham, 1968,
problem 17]
429. Nine stlcks are sufficient, the same number required to make a
pentagon.
[NCTM, 1978, p. 184]
430. Group the pennies in units of four, touching at the ends. To
allow the end pennies to touch three pennies, the ring must be
completed, which can be done with five groups, a total of twenty pen-
Illes.
[Kendall and Thomas, 1962, problem A18]
318 Penguin Book of Curious and Interesting Puzzles
431. 'Put your first and second fingers on 1 and 2, bring them round to
the corresponding position on the right-hand side. Then push the six
coins bodily to the left, leaving coins 1 and 2 in the position shown.'
[Abraham, 1933, problem 124]
432. This problem is presented by Mosteller, who records that John
von Neumann, a brilliant mathematician who was also a walking
computer, solved the problem 'in his head in 20 seconds in the
presence of some unfortunates who had laboured much longer'. If it
is any consolation - it probably won't be - to readers who find this
anecdote depressing, von Neumann also once solved the problem of
the fly between approaching trains (d. problem 288) by adding up an
infinite series in his head, rather than spotting that you only need to
divide the distance between the trains by the fly's speed.
Mosteller points out that some simplifying assumptions are needed
to solve this problem. It would be extremely complicated, and require
empirical testing, if for example the elasticity of the table were taken
into account. He suggests inscribing the coin in a sphere, as in the
figure, and assuming that the chance that the coin falls on its edge, is
the ratio of the portion of the surface of the sphere between the edges
of the coin to the total surface of the sphere. By the beautiful theorem
of Archimedes that was inscribed on his tomb, this zone will have one
third the area of the sphere when the coin's thickness is one third the
diameter of the sphere.
By Pythagoras, this thickness is about 35 per cent of the diameter
of the coin:
The Solutions 319
Edge
+
8
or -R' = r'
9
R' r'
9 8
1 .fi
-R = -r ~ 0.354 r
3 4
[Mosteller, 1987, problem 38]
433. Take a rectangular sheet of paper and use the ruler to draw two
parallel lines, as in the figure. Then placmg the ruler as indicated, the
angle BAC will be 30°.
[Wells, 1979, problem 58, part four]
320 Penguin Book of Curious and Interesting Puzzles
434. XY is the given segment. Draw two lines parallel to it, as in the
figure. Take any point A on the second line and join it to X and Y,
and then draw XB and YC to meet in o. Then the line AO passes
throught the mid-point of XV.
A
X ~ - - - - - - - - - - 4 - - - - - - - - - ~ ~ Y
Another solution, which involves placing opposite sides of the ruler
against opposite ends of the line segment, an operation sometimes
considered to be illegal, is illustrated in the figure below. The dotted
diagonal of the parallelogram bisects the original segment.
435. The piano will present the greatest difficulty when it is symmetric-
ally wedged into the corner. Two corners will then touch two sides of
the corridor, and the inner corner of the corridor will touch the
opposite side of the piano. The area of the piano being a maximum
when viewed vertically, this reduces to the problem of the largest
rectangle that can be cut from a triangle. In this case, the base of the
triangle is double its height, and so the piano has the same propor-
tions: it is twice as long as it is wide.
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
/
The Solutions 321
436. There are two other four-digit numbers with this property, 2025
and 9801, of which only 9801 has distinct digits.
437. 'Calling the unknown lengths of the scale-arms a and b, and the
parcel's true weight x ounces, by the principle of the lever
from which
and
bx = 28!a
ax = 36b
Xl = 1017
x = 31.89 ...
The parcel is therefore under two pounds and goes for sixpence. If
I had adopted the happy-go-lucky method of striking an average
between the recorded weights, I should have arrived at the figure of
32i oz, a weight demanding a ninepenny stamp.'
[Williams and Savage, n.d., problem 57]
438. One weighing IS enough. Take one coin from the first box, two
from the second, three from the third, and so on, up to ten coins from
the tenth box. Then the weight of all these coins will fall short of the
total if they were all sound, by 2 gm times the number of the box
containing the duds.
439. Three welghmgs are necessary. This is one schema:
In left-hand pan
2
2
4
3
3
7
10
11
10
4
7
3
In right-hand pan
5
8
6
6
9
9
10
12
322 Penguin Book of Curious and Interesting Puzzles
Denote a result in which the pans balance by 0, one in which the
right-hand pan falls by r, and one in which the left-hand pan falls by
I ... 'Interpretation of a weighings record is then a simple matter. If the
result is 000, all coins must then have the same weight. If the result is
rOl, we look for a coin with programme rOl, find this to be coin No.
4, and conclude that coin No.4 is too heavy. However, if the result is
air, for example, we find no coin with this programme, and conclude
that it is the coin with the opposite programme Orl - namely coin No.
7 - which then is too light. The extension to the analogous problem
for 39 coins with four weighings, or for 120 coins with five welghings,
and so on, is not difficult.
'This solution can be applied when thirteen coins are in question, if
it is known that exactly one of them is false. The weighings result 000
then indicates that coin No. 13 is false - a unique case in which there
is no indication whether the false coin is too heavy, or too light.'
[Sprague, 1963, problem 8]
440. Yes. Jack accepted Fred's bet and handed over two pounds.
Fred welshed on his bet, and handed over one pound, making a profit
of one pound.
441. This table shows the payoffs for Red, according to Black's
response. The only bids for Red which prevent Black from winning
are Sp and 6p, and of these 6p is the better: if Black foolishly bids
high, then Red will make a profit of 1p with his 6p bid, but would
break even by bidding Sp.
BLACK BID
3p 4p Sp 6p 7p 8p
3p 0 -2 -2 -2 -2 -2
4p 2 0 -1 -1 -1 -1
RED BID
Sp 2 1 0 0 0 0
6p 2 1 0 0 1 1
7p 2 1 0 -1 0 2
8p 2 1 0 -1 -2 0
PAYOFF to RED (p)
The Solutions 323
The result holds true regardless of the number of players (assuming
that the high bidders divide the pool and the liabilities equally).
Always bid one unit more than the stake.
[Silverman, 1971, problem 42]
442. Of the eight sandwiches contributed, each man ate 8/3 = 21
sandwiches. Therefore Jones contributed 21 of his sandwiches to
Watson's lunch, while Smith contributed only 1 of a sandwich, only t
as much. So Smith was right to insist that Jones was wrong, and the
£2 was actually divided into £1.75 for Jones and 25p for Smith.
443. Flowers cost lOp, whilst I.P.A. costs less than lOp. The first
stranger put down a lOp piece on the counter. The second man put
down a 5p piece, and five pence in copper coins.
[Kendall and Thomas, 1962, problem A9]
444. There are 8 x 8 = 64 individual squares, plus 7 x 7 = 49
squares composed of four individual squares, plus 6 x 6 = 36 squares
of nine individual squares ... and so on, making a grand total of:
8
2
+ 7
2
+ 6
2
+ 5' + 4
2
+ 3' + 22 + 12 = 204 squares.
445. This is one of the original constructions described by Lorenzo
Mascheroni in 1797 in his Geometria del Compasso, in which he
demonstrated that you do not need a ruler or straight edge to do
geometry: an ordinary pair of compasses will suffice by themselves.
He demonstrated this surprising fact by showing how all the basic
constructions of Euclidean geometry could be so performed, from
which it follows that all the more complex constructions based on
them are also solvable by compass only.
Draw two circles centred on the given points, A and B, and with
radius AB. From B, without changing the compass setting, mark off
/
/
I
I
\
\
\
~ - - - - - - - - - - - ~ - - - - ~ ~ ~ - - ~
CAE
B
324 Penguin Book of Curious and Interesting Puzzles
three arcs to find C, the opposite end of the diameter BAC. With
radius CB and centre C, draw an arc to intersect the circle centred on
B at D. FinaJly with radius AB and centre D, draw an arc which wiII
intersect AB at E, the mid-point of AB.
The isosceles triangles DBE and CDB have the same base angle at
B, and are therefore similar. Since DB is one half of BC, BE is one
half of BD = BA.
[Graham, 1963, problem 11)
446. One move. Pick up the third tumbler from the left and pour its
contents into the last tumbler.
[Always, 1965)
447. Yes, in six possible different ways. The eight arrangements of
heads and tails can very conveniently be displayed at the corners of a
cube, in such a way that each arrangement can be changed into an
adjacent arrangement by moving to an adjacent vertex of the cube.
__________

/////
//
HTT

There are three ways to set out from HHH on your journey, and
two choices of route after your first stop. Thereafter, the rest of the
route is forced. This is one solution:
HHH THH THT HHT HTT HTH TTH TTT
448. The two dates give 1 as first digit of 11 and 8 across.
15 (a cube) must be 27; for since 9 down ends in 1,64 would give a
3 for end of 7 d. (a square), which is impossible.
:.16 ac. is 16 and 9 d. is 11.
The Solutions 325
Since 10 d. = 10 ac. xII, 10 ac. must end in 2.
We can now calculate 14 ac. (perimeter of D.M.) to be 792.
Inspection of col. 2 shows that 12 d. = 19.
11 ac. must be 191" and 3 d. must begin with 1 or 2.
No. of roods being integral (see note), 1 ac. is an even multiple of
10, ... it ends 20.
Length and breadth of D.M. must be two numbers whose sum is
396 and product ends in 20, and as between them they contain the
factor 11 twice over, they can only be 220, 176.
It is now possible to fill in the diagram thus far:
'Shillings per acre' begins with 3, ... price per acre is between £15
and £20; and there are 8 acres .
... 4 d. is 142 and 1 d. is 355.
The 'down' number that = an 'across' number must be 10 d. And
11 times Farmer's age cannot = 352 .
... it is 792.
Finally, we want a square for 2 d., 7"*6. This can only be 84
2
or
86
2
• But 84
2
= 7056, which will not do for 5 ac.
".2 d. = 7396
and Mrs Grooby's age is 86.
449. Every row and every column must contain at least one black
square, so solutions with eighteen squares are easy and numerous.
The pattern on the left exploits the knight's move, to produce a
solution in seventeen squares, and no doubt for a sufficiently large
initial board this solution or its reflection will be maximal. However,
326 Penguin Book of Curious and Interesting Puzzles
by taking advantage of the limited size of the given board, a solution
in only sixteen squares is possible; as illustrated on the right.
[Wells, 1979, problem 63]
450. The solution is 242/303 = 0.7986798679867986 ...
Let F = .TALKTALK ...
Then 10000F = TALK.TALK... Subtracting,
9999F = TALK.
Then (EVE)/DID = F = (TALK)/9999. Therefore (TALK)/9999
when reduced to lowest terms equals (EVE)/DID, hence the denomina-
tor, 'DID' is a 3-digit factor of 9999, namely 101,303, or 909.
(a) Assume DID = 101. Then (EVE)/101 = (T ALK)/9999 or
TALK = (EVE)99 = (EVE)(100 - 1) = EVEOO - EVE. This leaves
an E as the first digit which therefore cannot be T. Therefore DID *"
101.
(b) Assume DID = 909. Then (EVE)/909 = (T ALK)/9999 or
TALK = (EVE)11 = (EVE)(10 + 1) = EVEO + EVE and this time
the units digit is E and therefore cannot be K. Therefore DID *" 909.
(e) Hence if there is a solution DID = 303. Since F is a proper
fraction, EVE can be only ( 1 ~ 1 ) or ( 2 ~ 2 ) , leaving for trial only 121;
141; 151; 161; 171; 181; 191; also 212; 242; 252; 262; 272; 282; 292.
All these except 242 repeat, in the quotient, a digit that appears in
the dividend so that finally 242/303 = .79867986 ... is the only solu-
tion.
[Beiler, 1964, p. 301, problem 66]
451. The number is 102564: .102564 x 4 = 410256. The answer is
easily found by starting with the fact that it ends in a 4:
The Solutions 327
4
and multiplying by 4, so that the next figure to the left is 6, with 1 to
carry:
64
Multiplying the 6 by 4 and adding the carry, the next figure is 5, with
2 to carry:
564
and, since (4 x 5) + 2 = 22, the next is 2 with 2 to carry:
2564
(2 x 4) + 2 = 10, so place 0 and carry 1:
02564
(0 x 4) + 1 1, so place the 1, with no carry:
102564
This is the solution.
452. The most likely total is 13. For every way in which you could end
up with a total of 14 or more, there is a way of ending up with 13: all you
have to do is to throw one less, and this is certainly possible because you
cannot first exceed 12 and arrive at a total of 14 or more by throwing a 1.
Therefore the number of ways of reaching 13 is at least as great as
the number of ways of reaching any higher total. But there are also
ways to first exceed 12 and reach 13, for example by starting With 12
and throwing a 1, which do not have any matching throws for higher
totals. Therefore there are more ways of getting to 13 and the
probability is greatest that your total will be 13.
[Honsberger, 1978, p. 42)
453. About 73 per cent more. The figure shows the shape of the
triangular cake and how it was cut. The original triangle and all its
parts have angles of 120
0
, 30
0
and 30". The ratio of a larger to a
smaller piece IS J3.
328 Penguin Book of Curious and Interesting Puzzles
454. Three pieces are sufficient. Here are two solutions.
I I
B B
C
C
A
A
B C
A
C
A
B
I
[Eastaway, 1982, p. 123]
455. Let the original cheque be for x dollars and y cents. Then
lOOy + x - 68 = 2(lOOx + y)
or 98y - 68 = 199x
If 98y - 68 is a multiple of 199, so will be its d o u b l ~ , 196y -- 132,
which is 3y - 63 short of 199y - 199.
This difference will be zero if y is 21, and this is the smallest
possible value for y. (It could also be 21 -+- any multiple of 199.) So
the original cheque was for $10.21.
[Beiler, 1964, p. 294, problem 10]
The Solutions 329
456. Ignoring reflections and rotations, this is the unique solution.
The original instructions did not forbid going outside the original
square, so that is clearly allowed.
457. This is the unique and elegant solution, again ignoring rotation,
which produces just one other solution.
[Schuh, 1968, p. 340)
330 Penguin Book of Curious and Interesting Puzzles
458. 'At nobody. Fire your pistol in the air, and you will have the
best chance among all three truellists!
'Certainly you don't want to shoot at Black. If you are unlucky
enough to hit him, Gray will polish you off on the next shot. Suppose
you aim at Gray and hit him. Then Black will have first shot against
you and his overall probability of winning the duel will be ~ , yours t.
Not too good. (The reader is invited to confirm Black's winning
probability of $ by summing the infinite geometric series: j + m m
m + m m m m m + ... )
'But if you deliberately miss, you will have first shot against either
Black or Gray on the next round. With probability j, Black will hit
Gray, and you will have an overall winning probability of ~ . With t
probability Black will miss Gray, in which case Gray will dispose of
his stronger opponent, Black, and your overall chance against Gray
will be t.
'Thus by shooting in the air, your probability of winning the truel
is M or about 40 per cent. Black's probability is fr or about 38 per
cent. And poor Gray's winning probability is only ; (about 22 per
cent).
'Is there a lesson in TRUEL which might have application in the
field of international relations?'
[Silverman, 1971, problem 79)
459. Let x and y be the times of arrival measured in fractions of an
hour between 5 o'clock and 6 o'clock. The shaded figure in this graph
shows the arrival times for which the duellists meet .
y
1
12
No meeting
.!.
12
~ ~ - - - - - - - - - - - - - - - - ~
o .!.
12
1
12
The Solutions 331
The area representing non-meeting is (11112)2 and so the area
representing meeting is 23/144, and the chance that they actually do
fight is a little less than 1/6.
[Mosteller, 1987, problem 26]
460. This is one solution. Numbering the matches 1-15 from left to
right: move 5-1, 6-1, 9-3, 10-3,8-14,7-14,4-2, 11-2, 13-15, 12-15.
The diagram shows the position after the first five moves of the solu-
tion.
~ ~
~ ~
~
~ ~ ~ ~ ~
~ ~ ~ ~ ~
2 3 4 5 6 7 8 9 10 11 12 13 14 15
[Kordemsky, 1972, problem 57]
461. Three moves are sufficient. Numbering the cups 1-7, in any
order you choose, you could invert 1,2 and 3; 3, 4 and 5; 3, 6 and 7.
If however, the rules specified that four had to be inverted at each
turn, then the problem is impossible, because after each inversion
there will always be an even number of cups the right way up, and
never an odd number, such as 7.
462. Here is a diagram of the twenty-five desks, shaded like a
chessboard. Notice that by following the teacher's instructions each
pupil moves to a desk of an opposite colour.
Peaky sat in one of the black desks, so on Tuesday there were
twelve pupils in black desks and twelve in white, and they could
follow the teacher's instructions, but when Peaky returned there were
thirteen pupils in black desks and only twelve in white, and so it was
332 Penguin Book of Curious and Interesting Puzzles
110.0.
0.0.0
110.0.
0.0.0
110.0.
not possible to follow instructions, which Peaky had the misfortune
to point out to the teacher.
[Adapted from NCTM, 1978, problem 115)
463. Since the problem rejects solutions which produce a square with
thickness, it must be intended to take the cube as its surface only. By
cutting along suitable edges the surface of a cube can be flattened into
one piece, in fact into a hexomino, in eleven different ways, all of
which can then be dissected into a square, though some require more
than four pieces.
This 'z' shape makes a cube and requires only four, by standard
methods. The top figure is composed of six identical squares. The
slanting line passes through its centre at approximately 54·7°. (Varia-
tion from this figure will produce a rectangle rather than a square.)
The Solutions 333
The second figure is composed of the two portions of the first figure
rearranged. The line through the centre of this figure is perpendicular
to the sloping edges. These pieces are then rearranged to form the
third figure.
[Games and Puzzles Journal, No.3, 1988, page 41]
464. With four vertical slices, and assuming that Jane's cake is not of
some extraordinary shape, a maximum of eleven pieces is possible.
If the cuts need not be vertical, then the first three cuts can create
eight pieces, for example by slicing twice vertically at right-angles,
and making the third cut horizontal. The fourth cut can then slice
through no less than seven of the eight pieces, making a total of
fifteen pieces. For example, take the plane cut which passes through
the mid-points of six edges and six eighth-cubes, and displace it
slightly.
465. Sunday.
[Abraham, 1933, problem 10]
466. Suppose that there are E errors in total and that the first proof-
reader, A, finds } of the errors, and the second, B, finds ~ of
them. Then ~ = 30 and ~ = 24.
Of the ~ that A found, B will have found ~ , and so from the errors
found by both
E
- = 20
xy
So the total expected number of errors IS (30 x 24)/20 = 36.
334 Penguin Book of Curious and Interesting Puzzles
Between them they found 30 + 24 - 20 = 34, so it is expected that
on average only two errors remain undetected by either of them.
467. Four. The sentence contains three spelling mistakes, plus the
false claim that it only contains one mistake, making a total of four
mistakes.
The second question, paradoxically, cannot be answered. It con-
tains only two spelling mistakes but claims to contain three mistakes;
therefore that claim is wrong and it actually contains three mistakes -
except that if it contains three mistakes then the claim that it contains
three mistakes is correct, and so it only contains the two spelling
mistakes, in which case ... !
468. 'Let n be the number of steps visible when the escalator is not
moving, and let a unit of time be the time it takes Professor Slapenar-
ski to walk down one step. If he walks down the down-moving
escalator in 50 steps, the n - 50 steps have gone out of sight in 50 units
of time. It takes him 125 steps to run up the same escalator, taking
five steps to everyone step before. In this trip, 125 - n steps have gone
out of sight in 125/5, or 25, units of time. Since the escalator can be
presumed to run at constant speed, we have the following linear
equation that readily yields a value for n of 100 steps:
n - 50 125 - n
50 25
[Gardner, 1966, Chapter 14, problem 4]
469. The best location is at X, on a vertical line which has one kiosk
to its left and one to the right, and on a horizontal line which has one
kiosk above and one kiosk below. Moving the kiosk to Y, for
example, would reduce the horizontal distance travelled by the owner
of one kiosk while increasing the distance travelled by two of them.
x
y
The Solutions 335
Of course, it cannot be denied that the criterion proposed, merely
to minimize the total distance travelled, has turned out to be unfair
on the owner of the north-east kiosk, who will travel much further
than either of the others - but that injustice was no part of the
puzzle.
[Sprague, 1963, probleml]
470. The chance of receiving a second Nobel prize only depends on
the total number of living Nobel-prize winners if you know for a fact
that the Nobel prize committee have decided to honour, again, one of
that select group. But Nobel prizes are awarded quite independently
of past awards, and so the actual chance of being awarded two Nobel
prizes is one chance in a billion times a billion (if you suppose, which
is implausible, that every human being on earth has an equal chance
of an award).
Linus Pauling's Nobel prizes, incidentally, were the prize for Chemis-
try and the Nobel Peace prize.
[Morris, 1970, problem 61. Morris does not give a satisfactory
solution]
471. The fallacy is exposed when you use the argument of the
problem to prove that any two horses are of the same colour.
Removing each of the horses in turn only leaves the other one horse,
and the set of N - 1 horses which are all of the same colour as each
other, and the same colour as the horses removed, has one member.
If only it were possible to conclude that any pair of horses were the
same colour, then it would indeed - and very obviously - follow that
any three, four, five, etc., horses were of the same colour.
472. The surgeon was the boy's mother.
473. His son.
474. The points are: A - 8
B-14
C- 9
The greatest number of roints that they can get for wins or draws
is: A - 5, B - to, C - 5 .... only two matches were played (not one, for
in that case one of them would have got no points).
Suppose that B drew two matches, against A and C, then B's total
of goals should equal (A + C)'s total. But B's total is 4, and A's is 3
and C's is 4.
336 Penguin Book of Curious and Interesting Puzzles
:. A v. C was a draw and B won one.
Either A or C played only one game, and since A got 3 goals and C
got 4, A v. C must have been 3-3. And C scored 1 goal against B.
:. B v. C was 4-1.
Complete solution
A v. C 3-3
B v. C 4-1
[Emmett, 1976, problem 82]
475. Supply four more identical triangles, and you will see that the
marked dots are on the diagonal of the completed square:
[Wells, 1987, p. to]
476. 9" is by convention interpreted as 9(") = 9387.420.48'.
This number, if written out without using index notation, contains
more than 360 million digits. Nevertheless, it is possible to find its
final two digits quite easily. The final digits of powers of 9 follow a
cycle, which can be found by multiplying the last two digits alone, by
9, again and again:
Power 0(9 Last d,gits
1 9
2 81
3 29
4 61
5 49
6 41
7
8
9
\0
11
69
21
89
01
09
The Solutions 337
Very conveniently, 9
10
ends in 01, and the cycle starts again. So
9387.420 .• 8., which IS 9" multiplied by a power of 9
10
, ends in the same
pair of digits as 9", that is, 89.
By a similar argument, based on the repeating sequence of the last
three digits, the last three digits of the power could also be calculated.
477. 'Among the five married couples no one shook more than eight
hands. Therefore if nine people each shake a different number of
hands, the numbers must be 0, 1, 2, 3, 4, 5, 6, 7, and 8. The person
who shook eight hands has to be married to whoever shook no hands
(otherwise he could have shaken only seven hands). Similarly, the
person who shook seven hands must be married to the person who
shook only one hand (the hand of the person who shook hands only
with the person who shook eight hands). The person who shook six
must be married to the person who shook two, and the person who
shook five must be married to the person who shook three. The only
person left, who shook hands with four, is my wife.'
[Gardner, 1977, p. 69, problem 5, contributed by Lars Bertil
Owe]
478. Between races, they agreed to swop horses, so that each was
then mounted on the other's horse, and determined to leave his own
horse trailing second.
479. 'Start the 7- and II-minute hourglasses when the egg is dropped
into the boiling water. When the sand stops running in the 7-glass,
turn it over. When the sand stops running in the II-glass, turn the 7-
glass again. When the sand stops again in the 7-glass, 15 minutes will
have elapsed.'
[Gardner, 1981, p. 190, problem 8.1, contributed by Karl Fulves]
480. This problem was first posed by T. P. Kirkman, a notable
amateur mathematician, in 1847, and repeated in the Lady's and
Gentleman's Diary for 1850. Only in 1969 did Ray-Chaudhuri and
Wilson show that there is a solution in the general case, in which the
number of girls is any odd multiple of 3.
338 Penguin Book of Curious and Interesting Puzzles
Sun. Mon. Tues. Wed. Thurs. Fri. Sat.
0,5,10 0, 1, 4 I, 2, 5 4, 5, 8 2, 4,10 4, 6,12 10,12,3
1,6,11 2, 3, 6 3, 4, 7 6, 7,10 3, 5,11 5, 7, \3 11,13,4
2,7,12 7, 8,11 8, 9,12 II, 12,
°
6, 8,14 8,10, I 14, 1,7
3,8, \3 9,10,13 10, II, 14 13,14, 2 7, 9,
°
9, II, 2 0, 2,8
4,9,14 12,14, 5 13, 0, 6 I, 3, 9 12,13, I 14, 0, 3 15, 6,9
This is one solution. Further discussion and references will be
found in Rouse Ball, 1974, p. 287.
481. Professor Sweet replied, in effect, 'Instead of three circles in a
plane, imagine three balls lying on a surface plate. Instead of drawing
tangents, imagine a cone wrapped around each pair of balls. The
apexes of the three cones will then lie on the surface plane. On top of
the balls lay another surface plate. It will rest on the three balls and
will be necessarily tangent to each of the three cones, and will contain
the apexes of the three cones. Thus the apexes of the three cones will
lie in both of the surface plates, hence they must lie in the intersection
of the two plates, which is of course a straight Ime.'
[Graham, 1963, problem 62]
482. This is the solution given by Harry Lindgren and Greg Frederick-
son, in their marvellous book Recreational Problems in Geometric
Dissections and How to Solve Them.
Since the final assembled star must have an edge length J3 times
that of the original stars, it is natural to join the vertices of a smaller
star to the centre, because that radius IS indeed J3 times the edge
length of the smaller stars.
[Lindgren and Frederickson, 1972, pp. 104-5]
483. The hidden feature is that the side of the square is equal in
length to the line joining a vertex of the dodecagon to the next-
vertex-but-three. Such simple relationships between polygons of equal
The Solutions 339
area invariably indicate that a mutual dissection is relative1y simple.
These are two solutions, each in six pieces. The first has the
remarkable property that it can be executed with no measutement at
all, simply by joining suitable vertices of the dodecagon.
[Lindgren and Frederickson, 1972, p. 41-2)
484. The smallest in area, and measured by the length of the shortest
side, is 3 x 7. This is one solution with attractive symmetry: the
knight starts at 1 and moves to 2, 3, .. . in sequence.
9
4
7
6
1
10
3
8
5
20
11
2
17
14
19
12
21
16
15
18
13
For the tour to have rotational symmetry, the board must be at
least 6 x 6. There are five such tours on the 6 x 6 board, of which
this is one:
[Kraitchik 1955, p. 263)
340 Penguin Book of Curious and Interesting Puzzles
485. The minimum solution requires seventeen moves.
fDomoryad, 1963, p. 118)
486. Right-handed.
487. There are eighty magic hexagons, in twelve of which the ourer
vertices also sum to the magic constant, 26.
[Gardner, 1978b, p. 58)
488. 'Eight swmgs are enough to reverse the two bookcases. One
solurion: (1) Swmg end B clockwise 90 degrees; (2) swing A clockwIse
The Solutions 341
30 degrees; (3) swing B counterclockwise 60 degrees; (4) swing A
clockwise 30 degrees; (5) swing B clockwise 90 degrees; (6) swing C
clockwise 60 degrees; (7) swing D counterclockwise 300 degrees; (8)
swing C clockwise 60 degrees.'
[Gardner, 1977, p. 180]
489. Slide the horizontal match to one side, by half its length, and
then move the unattached match to form the remaining side of the
cocktail glass, which now contains the cherry .
•
r
I I
, '
" II
II
II
II
II
II
II
II
II
II
U
490. Assemble the matches in this order, using long kitchen matches
in preference to the shorter kmd.
4 i
2===.:j
.3===
[Abraham, 1933, p. 58]
342 Penguin Book of Curious and Interesting Puzzles
491. Each symbol is one of the digits 1 to 9, with its reflection in a
vertical mirror. The n ~ m b e r s are arranged to form a magic 3 x 3
square, so the symbol in the empty cell is an 8 with its reflection.
[Plus, No.6, Summer 1987, p. 7)
492. The number of balls in a square pyramid is the sum of the layers
from top to bottom, which are the square numbers, 1,4,9, 16,25 ...
The number on the triangular pyramid is likewise the sum of the
triangular numbers, there being 1,3, 6, 10, 15, 21, 28 ... balls in each
layer from the top downwards.
To satisfy the conditions of the problem, a triangular pyramidal
number is one more than a square pyramidal number. By simple
addition, this first occurs when the square pyramid contains 55 balls
and the triangular pyramid, 56.
Since it does not occur again for any reasonably small sizes of the
two pyramids, this must be the solution.
(The formula for the number of balls in a square pyramid of side
n is in(n + 1)(2n + 1), and for the triangular pyramid, in(n +
l)(n +2).)
493. The sum of the areas of the given squares is 31152 which is not
a perfect square, but does equal 177 x 176, and it IS possible to
assemble them into the near-square shown opposite.
It is possible to assemble distinct squares into one large square, but
at least twenty-one pieces are required.
<0
,....
57
41
The Solutions 343
177
78
99
21
43
16
j
77
r-;5 34
494. Yes, you can, because one of the books has to contain no words
at all, and another contains just one word. If the number of books in
the library IS N, which is greater than the number of words in any of
them, the numbers of words must be N - 1, N - 2, N - 3 ... all
the way down to 3, 2, 1, O.
Since the problem states that there are 'books' in the library, you
can safely assume at least two books, but not more.
[Mensa, 1975, problem 27]
495. Extremely high. The number of molecules of air breathed by
Archimedes during his lifetime is vastly greater than the number of
litres of air in the earth's atmosphere. Assuming that the air breathed
by Archimedes has had plenty of time over more than 2000 years to
become thoroughly mixed with the whole atmosphere, and even
allowing for a proportion of the air which has been taken into the
ocean or used up in chemical reactions and remained locked away, it
is still highly likely that you are breathing some of Archimedes' air.
496. In half. This can be seen by a typical bit of argument by anal-
ogy.
Imagine 2 kg at A and 1 kg at each of Band C. For the purposes of
finding the centre of gravity of all three weights, the latter two can be
replaced by a 2 kg weight at A', and the centre of gravity will lie half
way between the two 2 kg weights, in other words, at Y.
Now think of the weights at A and C replaced by a 3 kg weight at
344 Penguin Book of Curious and Interesting Puzzles
2Kg
1 K g ~ - - - - - - - - - - ~ - - - - - - - - ~ 1 K g
B'. The centre of gravity now lies on the line B'B, dividing it in the
ratio, however, of 1: 3. So their intersection Y is indeed the centre of
gravity, and it divides AN in half and BB' in the ratio 3: 1.
497. Only one set, 1,2 and 3: 1 + 2 + 3 = 1 x 2 x 3 = 6.
498. He ties the rope round the tree on the shore, and then carries the
rope on a walk round the island. As he passes the halfway mark, the
rope starts to wrap around the tree on the island, and when he
reaches his starting point he ties the other end of the rope to the tree
on the shore and pulls himself across on the rope.
[Gardner, 1983, Chapter 8, problem 35]
499. This problem can be solved, appropriately, by use of the pigeon-
hole principle.
Dissect the field into four equilateral triangles, each with sIdes of
50 metres. The pigeon-hole principle says that if five objects are
placed in only four boxes, then one of the boxes contains at least two
objects.
In the present case, one of the triangular 'quarters' contains at least
The Solutions 345
two droppings, which will be at most 50 metres apart (allowing for
the possibility that one or more droppings might lie exactly on one of
the dividing lines).
500. Consider the coordinates of a pair of lattice points, and in
particular whether they are odd or even. The mid-point of the line
joinmg them will only be itself a lattice point if they match in parity.
But there are only four possible patterns of odd-and-even. A particular
point may have coordinates which are odd-odd, even-even, even-odd,
or odd-even.
Therefore if five points are taken, then by the pigeon-hole principle,
at least two of them must have the same pattern of coordinates, and
their mid-point will lie on the lattice.
[Larsen, 1983]
501.
502. If More starts, then Less has the perfect strategy of taking
enough matches to bring his take and More's last take up to the
magic number four. So if More takes two, Less takes two, but if
More takes three, then Less takes one, and so on.
This ensures that after five turns each, twenty matches will have
gone, and More loses by taking the last.
When it is Less's turn to start, he can only hope that More does
not know the winning strategy and will allow Less sooner or later to
take a number which brings the total taken up to a multiple of four.
If More does know the trick, then at least they will win alternate
games.
346 Penguin Book of Curious and Interesting Puzzles
503.
504. The maximum possible number with six lines is 20 which is just
the number of ways of choosing three lines out of six to be the sides
of the triangle.
505. These are probably the simplest solutions, and are easily
sketched on hexagonal or triangular paper:
[Gardner, 1989]
The Solutions 347
506. Angle BDE = 30°. Angle-chasing is not sufficient to find this
angle. A complicated general formula can be found by repeated use of
the sine rule, or, in this case, a simple geometrical construction can be
exploited.
A
Mark E' on AC so that E'BC = 20°. Then the three triangles EBC,
BE'C and DE'B are all isosceles. Therefore BEE' is equilateral, and
triangle EE'D is isosceles. But DE'E = 40°, and so BDE + 40° = 70°
and BDE = 30°.
[Tripp, 1975]
507. Suppose you start with coin 1. Count 1,2,3 and turn 4 tails up.
Count 6, 7, 8 and turn 1 tails up. Count 3, 4, 5 and turn 6 tails up.
Count 8, 1, 2 and turn 3 tails up. Count 5, 6, 7 and turn 8 tails up.
Count 2, 3, 4 and turn 5 tails up. Count 7, 8, 1 and turn 2 tails up.
508. 'Let N be the smallest integer. The product is then
N(N + I)(N + 2)(N + 3) = (N' + 3N)(N' + 3N + 2)
= (N' + 3N + I)' - 1
This is not a perfect square since two positive squares cannot differ
by 1.'
[Dunn, 1980, p. 92]
348 Penguin Book of Curious and Interesting Puzzles
509. 'Cleary 1 would not appear as a factor, and any 4 could be replaced
by two 2's, without decreasing the product. And if one of the factors
were greater than 4, replacing it by 2 and n - 2 would yield a larger
product. Thus the factors are all 2's and 3's. Moreover, not more than
two 2's are used, since the replacement of three 2's by two 3's would
increase the product. The largest number possible is therefore y2 x 2
2
"
[Dunn, 1980, p. 84)
510. 'He must pick up seven shirts to tide him over until the following
Monday. Hence he must deposit seven shirts each Monday. Counting
the shirt he wears on Monday, the required total is fifteen. (Note that
he cannot get by with only fourteen by exchanging his Monday shirt
for a clean one and turning it into the laundry, as he will be caught
short the following Monday.),
[Dunn, 1983, problem 84)
511. Because the overlap is bounded by two lines at right-angles
meeting at the centre of the larger square, it is equal to one quarter of
the larger square in area, or 25 square Inches.
512. A strip just 8 inches long is sufficient, as the diagram shows:
The strip is folded at 45° to bring it to the left (top figure) and then
back across the triangle, and so continues in a clockwise direction.
[Madachy, 1966, p. 124, problem 23)
The Solutions 349
513. The order makes no difference, the final price is the original
price multipled by (95/100) x (90/100) x (80/100), or 68.4 per cent.
514. The thickness will be 2
50
x rl; mm, which is rather more than
70,368,681 miles, 'or more than two-thirds of the distance from the
earth to the sun'.
[Tocquet, 1957, p. 109]
515. The simplest answer IS 9,876,543,210 - 0,123,456,789 =
9,753,086,421.
516. 99066
1
= 9,814,072,356.
517. Label the rectangle ABCD. Fold AC on to AB, td form the edge
CE, and then fold BD on to CE, to form the crease FG. Finally fold
BD towards A, so that the new crease passes through F and so that
BDA is a straight line.
F B
o
AB is then equal in side to the square whose area is equal to the
original rectangle. This is Abraham's proof:
'Let AB = x = y and AC = x - y. Then area of ABCD = Xl - yl
and AB - AC = 2y.
'First - Fold the short edge AC up on to the line AB as shown in
the first figure. CB is therefore equal to 2y.
'Second - Fold BD over to CE on line F and open out these folds.
BF is therefore equal to Y, and AF = x.
'Third - Fold DB about the point F, so that ADB is a straight line.
We now have the triangle AFB, and DBF is a right-angle. As AF = x,
BF = y, S is the side of a square which has an area equal to ABCD.'
[Abraham, 1933, problem 99]
350 Penguin Book of Curious and Interesting Puzzles
518. The pieces will only form a hollow triangle within a square:
'\,
',:::::::::::: ,:::::::::::::::::::::::::::::::::::::: : ::::::::::{:::
::::::::::: ::::: : : ~ : : : :::::: ::::::: ::: ' ,. :' :::::::: :::: ::::::::::
519. True. Label each person with the number of friends he or she
has at the party. If there are N people, the labels will be numbers
from 0 to N - 1. However, it is not possible for both 0 and N - 1 to
appear as labels, because if someone knows no one at the party, then
another person cannot know everyone. Therefore there are at most
N - 1 labels for N people and one of the labels must appear twice.
520. True. Consider anyone person at the dinner party (call this
person 'Tom' for convenience), and his relationships to the other five
present. Of these five, either at least three are friends of Tom, or at
least three are strangers to him.
If they include three friends of Tom, then either these three are
all strangers, or one pair are mutual friends and form with Tom a
group of three mutual friends. Similarly, if they include three strangers,
they are either mutual friends, or two of them and Tom are mutual
strangers.
521. In this diagram the intersections of the paths have been marked
with the numbers of ways in which Lady Merchant can reach the
intersection. For each intersection this number is the sum of the
numbers at the previous intersections from which the intersection can
be reached by walking along the side of one plot.
The Solutions 351
The numbers, in fact, are the numbers in Pascal's Triangle, and the
summer house can be reached in a total of seventy different ways.
522. Any triangular pattern of dots can be divided into a 'square'
(which does contain a square number of dots, although it is skew)
and two triangles.
o
o 0
If the 'square' mcludes the middle dot on one side, as in the middle
figure, then the next smaller square wiII leave two larger triangles
overlapping, as in the right-hand figure, and the addition of one extra
352 Penguin Book of Curious and Interesting Puzzles
dot allows the triangles to be separated. (If even smaller squares are
taken in this case, then the pair of equal triangles will overlap in a
larger triangle, and by adding that number to the original triangular
number, a similar dissection occurs.)
523. The figure shows one solution for each.
[Golomb, 1965, p. 26]
524. This is one solution:
[Golomb, 1965, p. 132]
525. These solutions were contributed by Bob Newman to Games
and Puzzles magazine. I have no proof that they are minimal, though
I suspect that they are.
On the left are individually locking 21-ominoes. In the middle are
14-ominoes which interlock as a tessellation, but not individually,
and on the right are 12-ominoes with the same property, but one half
of the tiles have to be turned over.
The Solutions 353
[Wells, 1979, problem 61]
526. 1,000,000,000 = 10" = 2" x 5" = 512 x 1953125.
Any power of 10 can be expressed as the product of the same
powers of 2 and 5, but the latter usually themselves contain at least
one zero. Two larger products which are zero-free are 10" and 10
33
•
[Ogilvy, 1966, p. 89]
527. For two children in general there are four equally likely events:
boy-boy, girl-girl, boy-girl and girl-boy. Since boy-boy is ruled out,
the chance of girl-girl is !.
Taking the same four events in older-younger order, both girl-boy
and girl-girl are ruled out, so the probability of two boys is t.
[Kordemsky, 1972, problem 236]
528. Mrs Tabako should place one cultured pearl in one jar, so that
if Mr Tabako chooses that jar his chance of success is 100 per cent,
and place all the remaimng pearls, forty-nine cultured and fifty
natural, in the other jar, so his chance will be 49/99.
[Morris, 1970, p. 136]
354 Penguin Book of Curious and Interesting Puzzles
529. Curiously, the answer depends on what means you choose to
define the random chord: the qualification 'random' by itself is not
sufficient to force a unique solution.
Consider first that the positions at which the chord meets the circle
are not important but the angle at which it meets the circle is, so
consider the chords through a given point on the circle. Comparing
their lengths with the equilateral triangle with a vertex at the same
point, the chord will be longer than the triangle-side if it falls within
the central angle of 60°, and the chance of this is 60/180 = !.
The Solutions 355
Next, consider the idea that the direction of the chord does not
matter - it is sufficient to consider all possible positions of the chord
parallel to a given direction. If the chord falls within the central band it
will be longer than the triangle side, otherwise not. But then length XY
is one half of the diameter of the circle, so in this case the probability is t.
A random chord can have its centre anywhere at all within the
circle, but only a chord whose centre lies within the circle inscribed in
the equilateral triangle will be longer than the triangle side. So the
required probability is the ratio of the areas of the smaller circle to
the larger, which is 1: 2', or i.
[Northrop, 1960, pp. 169-70. Other methods of interpreting the
qualification 'random' produce more different answers. Thus
Hunter and Madachy, 1963, p. 102, produce the
530. The figure can be divided into any number of identical pieces
(including two) by repeating the matching shapes of the ends.
[After Eastaway, 1982, p. 103]
356 Penguin Book of Curious and Interesting Puzzles
531. This shape is the only known pentagonal reptile of order 4, that
is, which divides into identical quarters.
532.
Divide the first figure about the vertical line of symmetry, and
bisect its other two tiles similarly, to get the second figure.
533. The process described is called Kaprekar's process, after D. R.
Kaprekar, an enthusiastic Indian amateur mathematician who has
been publishing his results in number theory for many years.
The result of the process, when sufficiently repeated, is always the
number 6174, called Kaprekar's constant. This is just one example,
starting with the number 4527:
7542 - 2457 = 5085
8550 - 0558 = 7992
9972 - 2799 = 7173
7731 - 1377 = 6354
6543 - 3456 = 3087
8730 - 0378 = 8352
8532 - 2358 = 6174
7641 - 1467 = 6174 and the calculation repeats.
534. 2 metres. The rollers themselves have moved forward 1 metre,
relative to the ground, so that the points at the top of the rollers on
which the slab originally rested are 1 metre ahead. The slab has
The Solutions 357
moved an equal distance relative to the rollers, making a total
movement of 2 metres.
[Northrop, 1960, p. 53]
535. He started at house No.5 and went 5-\0-1-9-2-8-3-7-4-6, a
total of 4900 metres.
[After Gardner, 1971, p. 235]
536. Neither. The distance between the heads does not change.
Think about it by imagining that only one bolt moves about the
other. It will either move towards the other or move away depending
on the direction of the screw on the bolt and the direction of rotation.
No matter what happens with the bolt you are holding, the
opposite will happen when you hold the first bolt still and move the
second around it. Therefore, when you 'twiddle' them, the two
motions will cancel each other out.
[Lukacs and Tarjan, 1982, p. 170]
537. No. Here is a counter example.
538. She takes eight stockings and is assured of at least one pair.
539. Extraordinary to relate, it is impossible to playa man as far as
the fifth rank, though any army of twenty men can send a man to the
fourth rank, to X, if the men are arranged as shown.
~
• • • • • • •
• • • • • •
• • • • •
• •
358 Penguin Book of Curious and Interesting Puzzles
Trial and error may well convince the reader of this impossibility,
which can be proved by the following beautiful argument, which
depends on picking a target square on the fifth rank, to which you
hope to despatch a man, and labelling it and surrounding squares as
in the figure below.
(ll
et
(li
0
2
0 1 0 0
2
0
3
0
4
0
5
0
6
0
5
0
4
0
3
0
2
0 0
2
0
3
0
4
0
5
0
6
0
7
0
6
0
5
0
4
0
3
0
2
0
3
0
4
0
5
0
6
0
7
0
8
0
7
0
6
0
5
0
4
0
3
0
4
0
5
0
6
0
7
0
8
0
9
0
8
0
7
0
6
0
5
0
4
0
5
0
6
0
7
0
8
0
9
eJ°
0
9
0
8
0
7
0
6
0
5
0
6
0
7
0
8
0
9
eJ°
eJ°
0
9
0
8
0
7
0
6
0
7
0
8
0
9
eJ°
eJ°
0
9
0
8
0
7
0
8
0
9
eJ°
eJ°
0
9
0
8
0
9
eJ°
eJ°
0
9
eJ°
eJ°
Here, the letter 0 stands for !(.j5 - 1), so that 0 has the property
that 0' + 0 = 1, and - this is the point of the labelling - every legal
move consisting of a solitaire jump leaves the sum total of the values
of all the squares occupied unchanged. For example, if a man occupy-
ing 0
7
jumps over a man occupying 0' it will be into a square
labelled 0
5
, and 0
7
+ 0' = 0
5
•
Now consider the total value of all the men in the initial army
below the starting line. This value must at least equal 1 if one man is
finally to reach square 1.
However, the sum of all the men in the first row below the starting
line is less than
(0
5
+ 0' + 0
7
+ ... ) + (0' + 0
7
+ 0" + ... )
0
5
0' 0
5
0'
=-- + --= -+ -= 0
3
+ 0
4
= 0
2
1-01-00' 0'
Similarly, the sum of all the men in the second row is less than 0
3
•
Continuing, the sum of all the men in the initial army is less than
The Solutions 359
0
2
+0
3
+0
4
+ ...
0
2
1 - 0
Therefore no man can reach the fifth rank. If, however, just one cell,
no matter where, is allowed to be occupied by two men, then the fifth
rank is within reach.
[Beasley, 1989, pp. 86-7. John Beasley, who has also written
The Ins and Outs of Peg Solitaire, comments that 'Solitaire
offers many lovely problems, but this is one of the loveliest']
540. Yes, it is. Here is a square surrounding seventeen points.
The general answer is 'Yes' also. An intuitive proof can start with
the idea that if you have a square which surrounds exactly N - 1
points, then by keeping the same centre and orientation you can
increase the size of the square until either one point lies on its
perimeter, in which case a sufficiently small further increase will
bring that point inside without putting any more points on the
perimeter, or several points appear on the perimeter at once, in which
case you retreat slightly, give the square a very slight movement, and
increase the size again until this time just one new point appears on
the perimeter.
[Honsberger, 1973, p. 121, where Honsberger gives Browkin's
watertight proof of this conclusion]
541. James's book was published in 1907, two years after Einstein
published his first paper on his special theory of relativity, so it is
appropriate that the answer is relative - it depends what you mean by
the verb 'circle'.
It is a well-known fact that the moon always turns its same face to
us, and that its distance from the earth was once less than it is now.
360 Penguin Book of Curious and Interesting Puzzles
We can imagine, therefore, a planet with a moon which always faces
it, and which circles it in phase with the planet's own rotation. Such a
moon would always face the planet at the same point in the sky.
Relative to the position of the planet in space, the moon is circling
the planet, and the planet is also circling the moon, relative to the
moon's path through space, but the planet and moon are never
behind each other because of their own matching rotations. Therefore,
if your idea of circling includes the idea of 'going behind' then they
do not circle each other.
The situation of the hunter and the squirrel is basically the same,
except that the tree does not move through space. Relatively simple?
542. 12128 farthings = £12 12s 8d, or £12112/8.
543. There are eight possible sets of crossings at the three intersections
and only two of these create a knot, so the probability is 114.
544. 72 = 9 x 8 so the amount is divisible by 8 and 9. Thousands
are always divisible by 8, so 79- is a multiple of 8, and so the last
digit is 2.
The sum of the digits is divisible by 9, since the number is a
multiple of 9, and so the first digit is 3. The invoice was for £367.92
and each bird cost £5.11.
[Beiler, 1964, p. 302, problem 73]
545. 'On the first day, the chauffeur was spared a 20-minute drive.
Thus Mr Smith must have been picked up at a point which is a 10-
minute drive (one way) from the station. Had the chauffeur proceeded
as usual, he would have arrived at the station at exactly 5 o'clock.
The 10-minute saving means that he must have picked up Smith at
4.50. Thus Smith took 50 minutes to walk what the chauffeur would
take 10 minutes to drive. From this we see that the chauffeur goes five
times as quickly as Smith.
'Now, on the second day, suppose that Smith walks for 5t minutes.
The distance he covers, then, would take the chauffeur only t minutes
to drive. Accordingly, Smith was picked up this time at t minutes
before 5 o'clock, that is, at 60 - t minutes after 4 o'clock. However,
starting at 4.30 and walking for 5t minutes, Smith must have been
picked up at 30 + 5t minutes after 4 o'clock. Hence 30 + 5t = 60 -
t, and t = 5. Therefore the chauffeur was spared a 5-minute drive
(each way), providing a saving of 10 minutes this time.'
[Honsberger, 1978, problem 6]
The Solutions 361
546. The train takes 30 seconds to travel 1 km, plus 3 seconds for the
complete train to pass any point, making a total of 33 seconds.
547. The paddle, being unpowered, does not move relative to the
water, and since the boat has a constant speed relative to the water,
regardless of that speed, it will take the boat as long to get back to
the paddle - 10 minutes - as it did .to get away from it. Since the
paddle (and the water) moved 1 mile in that total period of 20
minutes, the current must have been 3 mph.
[Graham, 1968, problem 24]
548. It is not true that if one sees two sides the other will see three. In
most positions, they will each see two sides. A quick sketch will show
that the chance of seeing three sides at once is rather small, and only
approaches fifty-fifty as the spy recedes to an infinite distance from
the building.
549. Let the longer candle be x inches originally, and burn at r inches
per hour. Then the shorter candle was originally x-I inches long.
Let s denote its rate of burning. At 8.30, the longer candle has burned
for 4 hours and the shorter for 2t hours, and they are the same length:
5s
x - 4r = (x - 1) - -
2
Also, the longer candle is consumed in 6 hours and the shorter in 4
hours, so:
6r = x and 4s = x-I
From these equations, r = It, s = 2 and x = 9, so the candles
were 9 inches and 8 inches long.
[NCTM, 1965, problem 76]
550. The second box from the right was originally in the middle.
'If the boxes are thought of as dice with numbers on them (convention-
ally arranged so that pairs of numbers on opposite faces add up to 7)
the sum of the numbers for the upper, front and right-hand faces of
any of the dice is an odd or even number according to its orientation;
for instance the diagram below (assuming the face with the A has one
pip on it) we have
362 Penguin Book of Curious and Interesting Puzzles
1 + 2 + 3 = 6 (even)
2 + 6 + 3 = 11 (odd)
4 + 2 + 1 = 7 (odd)
In this way we can call the orientation of a box either odd or even.
The class of orientation changes each time a box is tipped over on an
edge. This diagram shows this for two cases - where the first box is
tipped over backwards, or to the right. The result follows in the
general case from the fact that each tipping over of the box leaves
two of the previous numbers to be considered again, while the third is
replaced by its difference from 7 and thus becomes odd if it was even
before, and conversely. So the sum of the upper, front and right-hand
faces changes in similar fashion - the box goes from an odd orienta-
tion to an even one, or vice versa. It is not in fact difficult to show
that successive tipping operations can indeed produce any even orienta-
tion on any square of unchanged colour and any odd orientation on
any square of altered colour.
o
o
o
o
o
o
o
o 0
o 00
o
o
o
'If all the boxes in position A are taken to have even orientations,
only the second from the left in position B has an odd one; so the
latter must have been tipped over an odd number of urnes, and the
rest an even number of times.
'Now we assume that all this takes place on a floor which is
chequered like a chessboard, where each box exactly covers one
square. Every time a box is tipped over, this changes the colour of the
square which it covers; so each box stands on one colour in all its
even orientations, and on the other colour in all its odd ones. Suppose
that the first, third and fifth boxes are on black squares in position B;
then they were on black squares in position A also. The second box
in position B, which is now on a white square, must in view of its
new odd orientation have been on a black square, earlier, in position
A. So we recognize these four boxes as those which occupied four
like-coloured squares in position A - i.e. the outer ones. The second
box from the right in position B, with an even orientation on a white
The Solutions 363
square, stood also on a white square in position A, and so was in the
middle.'
[Sprague, 1963, problem 3]
551. Call the trains A and B. A runs one third of its length in 1
second and B runs one quarter of its length, so they separate by 1 +
1 = n of the length of either in 1 second. To pass each other, they
must separate by twice the length of either, which will take 2 x ¥ =
¥ = 3 ~ seconds.
[Workman, 1920, p. 406, exercise 7]
552. 4
3
= 64. In general, if two sides of the original triangle are
divided into N parts, there are N3 triangles in the figure, made up of
Nl triangles whose sides include the base of the original triangle, and
2 x tN(N - 1) x N whose sides do not.
s
A c
B
The answer can also be found by visualization: break the base of
the triangle, and slightly separate the lines going into each corner.
Continue to bend the two halves of the base and to separate the same
lines until the figure is transformed into a grid of squares. Each
original triangle to be counted corresponds to a rectangle, at least one
of whose sides lies along the edges AB and Be. The number of such
rectangles is the total number of rectangles in the figure, less the
364 Penguin Book of Curious and Interesting Puzzles
rectangles in the top square PQRS. The first number of rectangles is
T'N where TN is the Nth triangle number, tN(N - 1), and the second
number is P
N
_ I' The differences between the squares of the triangular
numbers, are the cubes of the integers: T'N - T'N _ I = N3.
[Wells, 1983--6, Series 1, problem 28]
553. He asks either of the warders, 'Does the warder who is guarding
the road that leads to freedom tell the truth?' If the warder he asks
replies 'Yes', he goes through that warder's door. If the warder
replies 'No', he walks to freedom through the other door.
554. Consider the puzzle with statements 1-5 only, and denote '2 is
true' by 2T, and so on.
2T implies ST implies 4F (since 4T, ST would make 4F).
4F implies two consecutive true statements, which is impossible.
Therefore 2F. 2F implies 4T (since 4F would have to be true),
which implies 3F and SF. Thus IF 2F 3F 4T SF, and the answer to
this shortened puzzle would be '4 alone'.
But adding the statement 6, if 6T then its removal would affect the
answer, and therefore 6F. Thus since 6F, the answer cannot be 4
alone. 2T can again be ruled out making 2F. Hence again 4T so 3F
and SF. The only combination left which works and does not give the
answer '4 alone' is IT 2F 3F 4T SF 6F. Statements 1 and 4 are true.
[Eastaway, 1982, p. 15]
555. Imagine a sphere inscribed in both the cylinders, and therefore
inscribed also in their common volume. Take any vertical slice
parallel to the axes of the two cylinders. The slice will show a circle,
which is a cross-section of the inscribed sphere, and a square circum-
scribed about it.
Add up all the circular slices and you will get the volume of the
inscribed sphere, which is ~ 1 t if the radius of each cylinder is taken to
be 1 unit.
Add up all the square slices and you get the volume of the solid
common to both cylinders. But each square slice has the same ratio to
its inscribed circle, 4 to 1t, and so the volume of the common solid has
the same ratio to the volume of the inscribed sphere.
The common solid therefore has volume If. It is notable that it is a
rational number, and does not involve 1t.
[Gardner, 1977, p. 176]
The Solutions 365
556. The four points of intersection of circle E, which needs no
proof, and the points A, B, C and D.
[Barr, 1969, problem 51]
557. Three boys and three girls each received one :1 cent and two 1
cent buns. There could also have been only one boy and one girl, but
then the purchase could have been made in ten ways instead of the
single way required.
[Beiler, 1964, p. 298, problem 32]
558. Black can only have moved his knights, which are back on their
original squares (having possibly swopped places), and his rooks
which can only have moved to the original position of a knight and
back again, and so Black has made an even number of moves.
Similarly White's knights and rooks have only made an even number
of moves. Therefore, to account for the odd single move by the White
pawn, the White King or Queen must have made an odd number of
moves between them, and this is most briefly accomplished by the
White King moving seven times, for example Ke1-f2-e3-f4-g4-g3-
366 Penguin Book of Curious and Interesting Puzzles
f2-el, in which the move to g4 ensures the odd number of moves
before return to this starting square.
So each player has made at least eight moves.
[Beasley, 1989, p. 77]
559. 'Before leaving the house I started my clock, without bothering
to set it, and I noted the exact moment A of my departure according
to its reading. At my friend's house I noted the exact times, hand k,
of my arrival and departure by his clock. On returning I noted the
time B of my arrival according to my clock. The length of my absence
was B - A. Of that time k - h minutes were spent with my friend, so
the time spent in travelling, t, in each direction, was 2t = (B - A)
(k - h). Thus the correct time when I reached home was k + t.'
[Kraitchik, 1955, p. 39, problem 49]
560. Tom lived at No. 81. The diagram shows the successive groups
of numbers that emerge from the answers to the first three questions.
Consider the last question that John asks. There is only one group
(16, 36) that would enable John to determine the number, whether
the answer had been YES or NO. The answers that John received
therefore must have been (in order) NO, YES, NO. Since Tom lied to
the first two questions, the correct answers should have been YES, NO,
YES, leading to the unique answer of 81.
8-100
greater than 50
YES NO
51,52 .. 8,9 ..
.. 99,100 .. 48,49
multipleof4
1
52,56.. 51,53 ..

David Wells was born in 1940. He had the rare distinction of being a Cambridge scholar in mathematics and failing his degree. He subsequently trained as a teacher and, after working on computers and teaching machines, taught mathematics and sCIence in a primary school and mathematics in secondary schools. He is still involved with education through writing and working with teachers. While at university he became British under-21 chess champion, and in the mIddle seventies was a game inventor, devising 'Guerilla' and 'Checkpoint Danger', a puzzle composer, and the puzzle editor of Games & Puzzles magazine. From 1981 to 1983 he published The Problem Solver, a magazine of mathematical problems for secondary pupils. He has published several books of problems and popular mathematics, including Can You Solve These? and Hidden Connections, Double Meanmgs, and also Russia and England, and the Transformations of European Culture. He has written The Penguin Dictionary of Curious and Interesting Numbers and The Penguin Dictionary of Curious and Interesting Geometry, and is currently writing a book on the nature, learning and teaching of mathematics.

Ihis copyrighl p.ge The mor.1 righl of Ihe aUlhor h". been ","erl<-d Typesel by OarlX Inlernalion.1 LImned, Bung.y, Suffolk Filmsel in Monopholo Sa bon Primed in England by Clay, LId, SI Ives pic Excepl in Ihe Unlled Slales of America, Ihls book IS ,old sublecl 10 Ihe condilion Ih.1 il shall nOI, by way of lrade or olher"i,e, be lenl, re-sold, hired OUI, or olherwlse circulaled wilhoul Ihe publisher's
prior consent in any form of binding or cover other than th..tt
In

which il is published .nd wilhoU[ • SImilar condilion including Ihi, condilion being Imposed on Ihe subsequenl purchaser

Contents

Acknowledgements Introduction Vll
The Puzzles The Solutions 179

VI

Bibliography Index 379

373

Acknowledgements

Please note that detailed sources for puzzles are given at the end of each puzzle solution, where appropriate. Grateful acknowledgement is given to the following: Dover Publications, Inc., for permission to reproduce material from: Stephen Barr, Second Miscellany of Puzzles (1969); A. H. Beiler, Recreations in the Theory of Numbers (1966); Angela Dunn (ed.), Mathematical Bafflers (1980), and The Second Book of Mathematical Bafflers (1983); L. A. Graham, Ingenious Mathematical Problems and Methods (1959), and The Surprise Attack in Mathematical Problems (1968); J. A. H. Hunter, More Fun with Figures (1966); F. Mosteller, Fifty Challenging Problems in Probability (1987); F. Schuh, The Master Book of Mathematical RecreatIons (1968); George J. Summers, New Puzzles in Logical Deduction (1968). I will also note here that although the original Loyd and Dudeney books are long out of print, two collections of Loyd's puzzles, both edited by Martin Gardner, are pu blished by Dover under the titles Mathematical Puzzles of Sam Loyd and More Mathematical Puzzles of Sam Loyd, and they have also reprinted Dudeney's Amusements in Mathematics. Robert Hale Ltd, for permission to reproduce 'Room for More Inside', from Gyles Brandreth, The Complete Puzzler (1982). McGraw-Hill, Inc., for permission to reproduce 'The True!', from David Silverman, Your Move (1971). Weidenfeld and Nicholson for permission to reproduce the Tangram puzzles from E. Cuthwellis (ed.), Lewis Carroll's Bedside Book (1979). John Hadley for the translation of Alcuin's Propositiones ad acuendos juvenes, and David Singmaster for lending me his copy, as well as giving me the run of his library of mathematical recreations. John Hadley's complete translation has subsequently been published in the Mathematical Gazette, Vol. 76, No. 475, March 1992. Finally, I should like to thank the staff of the British Library for their courteous help.

I hope that in due course they will form a separate volume. that this is not a history. and other famous puzzlers of that era such as Lewis Carroll and Eduard Lucas. One day a history of puzzles will be written. only. A boundary also had to be drawn between puzzles of a logical and mathematical nature. but generally speaking problems which require any mathematics beyond the most elementary algebra and geometry. up to the nineteenth century. followed by rapid expansion in the nineteenth century. This book follows that pattern. A number of puzzles are included which relate to mathematical recreations or which led to the development of specific recreations. Limitations of space have forced a strict selection. ThiS boundary cannot be drawn precisely. who has spent many years delving into the origins of popular puzzles. so it is no surprise that puzzling problems are as old as history itself. but the recreations themselves are not treated. Word puzzles are entirely excluded. and few of the puzzles require even that level of sophistication. who straddle the nineteenth and twentieth centuries. and mathematical recreations and mathematics itself. The second half of the book is devoted to the great variety of puzzles composed in the twentieth century. I hope by David Singmaster.Introduction
To puzzle and be puzzled are enjoyable experiences. have been excluded. and an explosion in the twentieth. These are followed by examples of the puzzles of Loyd and Dudeney. I have merely selected some representative figures. The first third is devoted to puzzles from the dawn of history.
. Readers interested in mathematical recreations will find references to many of the bestknown and most readily available sources in the bibliography. well justified by their immense richness and variety. of the puzzling questions that have found popular favour over the centuries. but in the meantime this book will give readers examples. in Egypt and Babylon. I must emphasize. and follow a similar pattern of many centuries of slow progress. however.

manipulative puzzles requiring some kind of apparatus also deserve a book-length treatment of their own. I hope that readers will find some of that pleasure in the immense variety of puzzles assembled here. and work was a pleasure hard to distinguish from play. and are excluded here.V11l
Introduction
Finally. I shall be happy to receive readers' opinions and suggestions. All the puzzles in this book can be tackled either mentally.W.
1992
. Compiling this book has taken me back to the days when I was Puzzle Editor of Games & Puzzles magazine. I wish you happy and successful puzzling!
D. or with the assistance of at most pencil and paper and perhaps a few counters. though I cannot guarantee to respond to every letter personally.

The Puzzles
.

.

in each sack there are seven loaves of bread. freely paraphrased here.The World's Oldest Puzzle
1. It dates from about 1650 BC. in each loaf there are seven knives. about 2800 years after Ahmes. The scribe's name was Ahmes. There are seven houses each containing seven cats. so the original of the Rhind papyrus was written in the same period as another famous source of Egyptian mathematics. and he states that he is copying a work written two centuries earlier. Fibonacci in his Liber Abaci (1202) posed this puzzle: 2. dating from 1850 BC. There is an undoubted fascination with geometrical series. Each cat kills seven mice and each mouse would have eaten seven ears of spelt. Seven old women are travelling to Rome. along an historical path that we can no longer trace. Henry Rhind. which is named after the Scottish Egyptologist A. What is the total of all these? This puzzle. and each knife has seven sheaths. and the number 7 is not only as
. is problem 79 in the Rhind papyrus. our richest source for ancient Egyptian mathematics. written on both sides. who purchased it in 1858 in Luxor. Returning to the cats and mice. On each mule there are seven sacks. of the Rhind puzzle? Not necessarily. the Moscow papyrus. The resemblance is so strong that surely Fibonacci's problem is a direct descendant. The question is to find the total of all of them. The Rhind papyrus is in the form of a scroll about eighteen and a half feet long and thirteen inches wide. and each has seven mules. Each ear of spelt would have produced seven hekats of grain.

and you naturally arrive at two similar puzzles.
The St Ives Riddle
3. The only fractions they used were i and the reciprocals of the integers. where he described what is now called the greedy algorithm. + ~. The Rhind papyrus contains a table of fractions in the form 21n for all odd values of n from 5 to 101. produces a sequence of no more than p unit fractions. Every cat had seven kits. but was especially easy for the Egyptians to handle. without repetition? Yes. Curious though this treatment of fractions may seem to us. no doubt it seemed both natural and easy to them. Subtract the largest possible unit fraction. but with one remarkable peculiarity. Can all proper fractions be expressed as the sum of unit fractions. and so on. How many were going to St Ives?
This rhyme appears in the eighteenth-century Mother Goose collection. Similarly. Thus their answer to the problem. They also had a rule for expressing i of a unit fraction as the sum of unit fractions: to find i of t. 'divide seven loaves among ten men' was not 7/10 of a loaf each. as Fibonacci showed.4
Penguin Book of Curious and Interesting Puzzles
magical and mysterious as any number can be. Put these factors together. i of i is -k + is. and 7 = 1 + 2 + 4. also in his Liber Abaci. cats. Is it also descended from the Rhind papyrus and Fibonacci?
Egyptian Fractions
The Egyptians could easily handle simple fractions. where p is less than q. Every sack had seven cats. the so-called unit fractions with unit numerators. The greedy algorithm does not work so well if we add the
. I met a man with seven wives. Kits. 4. but the fraction t + t. Every wife had seven sacks. sacks and wives. because they multiplied by repeated doubling. then do the same again. multiply 5 by 2 and by 6: i of t = rl.
As I was going to St Ives. Sylvester proved in 1880 that applying this greedy algorithm to the fraction plq.

in effect. How can 1 be represented as the sum of unit fractions with square denominators. 'I think of a number. 'If the scribe says to thee. while arithmetic progressions just plod along. Yet puzzles about arithmetical pr->gressions can be thoughtprovoking.the number?' Readers will naturally wish to express the answer in Egyptian fractions!
Sharing the Loaves
Arithmetic progressions have not been as popular in the history of puzzles as geometric ones. Which has the smallest largest denominator?
5. What is the smallest fraction 3/n for which the greedy algorithm produces a sum in three terms. but two terms are actually sufficient?
The sum of the series 1 + 112' + 113 ' + 114' . 'A number. step by equal step. What was the number?' 9. Two-thirds of the number plus its tenth make to.
. What number did I think of?' 8. There is after all something impressive. for example. makes 37. My answer is 10. 1. 6. = x'/6. mysterious even. There are just five ways to represent 1 as the sum of the smallest possible number of Egyptian fractions. 'Twothirds is to be added. and add to it two-thirds of the number. "10 has become j + to of what?'" is the Egyptian way of saying. Problem 29 of the Rhind papyrus is not quite so clear. There remains to. but might equal. so the sum of different Egyptian fractions whose denominators are squares cannot exceed x'/6.. It reads. in the rapidity with which geometric progressions increase. What is . I then subtract one-third of the sum. but it is plausibly the first ever 'Think of a Number' problem.' In clearer language that reads: 'I think of a number. and plus its half.The Puzzles
5
condition that all the denominators must be odd. plus its seventh. as this example illustrates. with no denominator greater than 35 ' ?
Think of a Number
7.. with odd denominators. One-third is to be subtracted. plus its two-thirds.

' The meaning is: 'Divide 100 loaves between five men so that the shares are in arithmetical progression. and so on.30 and the area is 11. The length of the marked side is 6. Instead of counting in tens and hundreds and using tenths and hundredths.30.30 means 6 + (30/60). or 11i. The Babylonians counted in a sexagesimal system.'
The Babylonians
Babylonian mathematics was arithmetical and algebraic and far in advance of Egyptian mathematics of the same period. or 61. based on a suggestion by the historian Moritz Cantor that the Egyptians might just possibly have made right-angles this way. so 6. 'A hundred loaves to five men.30 means 11 + (22/60) + (30/3600). Other problems were far more advanced. one-seventh of the three first men to the two last. The side of one is f + i the side of the other. This 'fact' is actually a myth.6
Penguin Book of Curious and Interesting Puzzles
10. which required the construction of an arithmetical series to fit given conditions. or that they had any knowledge whatsoever of Pythagoras's theorem. contempor-
.. however. consider problems about areas and square numbers. There is no evidence that they did anything of the sort. the area of a square of 100 is equal to that of two smaller squares. A triangular field is to be divided between six brothers by equidistant lines parallel to one side. and the sum of the two smaller shares is one-seventh of the sum of the three greatest. This is from the Berlin papyrus: 11. and 11..22. they used multiples of 60. What is the difference between the brothers' shares?
This problem is much like Problem to. They could solve all the problems in the Rhind papyrus and many more besides.22. Let me know the sides of the two unknown squares. 'If it is said to thee .'
Squares Without Pythagoras It is a well-known 'fact' that the ancient Egyptians used knotted ropes to make a 3-4-5 triangle and hence construct accurate right-angles. Thus a tablet from about 1600 Be.
Dividing a Field
12. They did.

to
the present day. A ladder of length 0.
The Greeks
Archimedes' Cattle Problem Archimedes (287-212 Be) was the greatest mathematician of antiquity. unlike the Egyptians. leads in modern notation to the solution of two equations of the form: bx' cy'
xy=a
-+-+d=O y x
which leads to an equation in
x·. consisting of the sum of two squares. This is from about 1800 Be: 'An area A.6.89 in the last.42 in the first line to 106/56 = 1. invented hydro-
. not only knew Pythagoras's theorem. What are the sides of the squares?' Pythagorean Triples The Babylonians. Problem: find the hypotenuse and one leg of a right-angled triangle whose ratio is approximately 1. how far will the lower end move out from the wall? 'Plimpton 322' is the name of a clay tablet dating from between 1900 Be and 1600 Be. Their investigations of Pythagorean triples started a trail of discovery. If the upper end slides down the wall a distance of 0. Although the lengths given seem to vary in an apparently irregular way from one line to the next. leading through Diophantus
to
Fermat. 15.
13. but they were also familiar with Pythagorean triples.The Puzzles
7
ary with the Rhind papyrus.54. It contains fifteen numbered lines with two figures in each line which are the hypotenuse and one leg of a right-angled triangle. Ladders were a natural source of problems. is 1000. triples of whole numbers such as 3-4-5 which are the sides of right-angled triangles. The side of one square is 10 less than two-thirds of the other square.
14. in fact their ratios increase steadily from 169/119 = 1. a wonderful geometer who anticipated the calculus. x' and a constant.30 is standing upright against a wall.

In each herd were bulls. another glossy black'. the dappled. giving separately the numbe~ of well-fed bulls and again the number of females according to each colour. the whole of the yellow. when the yellow and the dappled bulls were gathered into one herd they stood in such a manner that their number. who once upon a time grazed on the fields of the Thrinician isle of Sicily. 'If thou art able. and the plains of Thrinacia. they stood firm. mighty in number according to these proportions: understand. understand also all these conditions reg~rding the cows of the Sun. that the white bulls were equal to a half and a third of the black together with the whole of the yellow. compute the number of cattle of the Sun. beginning from one. 0 stranger. stranger.' Archimedes' cattle problem is extant in more than one manuscript. together with all the yellow. while the black were equal to the fourth part once more of the dappled and with it a fifth part. went to pasture together. and studied giant numbers in his book The Sandreckoner. When the white bulls mingled their number with the black. to find out all these things and gather them together in your mind.. thou wouldst not be called unskilled or ignorant of numbers. giving all the relations. the third yellow and . once more. one milk white.
. stretching far in all ways. Now the dappled in four parts were equal in number to a fifth part and a sixth of the yellow herd. grew slowly greater till it completed a triangular figure. while the black were equal to the fourth part of the dappled and a fifth. the number of cattle of the Sun. These were the proportions of the cows: the white were precisely equal to the third part and a fourth of the whole herd of the black. but not yet shalt thou be numbered among the wise . Observe further that the remaining bulls. together with. 'But come.the last dappled. were equal to a sixth part of the white and a seventh. It is a curiosity that one extremely difficult problem and one simple recreation are associated with his name. Again.. equal in depth and breadth. If thou canst accurately tell. Finally the yellow were in number equal to a sixth part and seventh of the white herd. thou shalt depart crowned with glory and knowing that thou hast been adjudged perfect in this species of wisdom. were filled with their multitude. including the bulls. there being no bulls of other colours in their midst nor none of them lacking. 16. divided into four herds of different colours. 'If thou art diligent and wise. 0 stranger. when all. 0 stranger.8
Penguin Book of Curious and Interesting Puzzles
statics.

and Archimedes' interest in very large numbers is evident from his Sandreckoner. A. and the reader is willing to be judged merely 'not unskilled' in the art. Archimedes dedicated the problem to his friend the great Alexandrian astronomer Eratosthenes.
Loculus of Archimedes
Several ancient sources refer to this puzzle. rather than perfectly wise.545 digits. If this is so. the latter conditions are ignored. however. such was his fame. which suggests that it was extremely difficult. Further details will be found in Sir Thomas Heath's A History of Greek Mathematics. in which he calculated the number of grains of sand needed to fill a sphere whose centre was the centre of the earth and which extended to reach the sun. The Book of Archimedes on the Division of the Figure Stomaschion. which is described in an Arabic manuscript. in classical antiquity a difficult problem was often described as a problema bovinum or a problema Archimedis.
. it could be merely a rectangular number. If. then the answer will be found in the Solutions section. 319. These conditions are ambiguous: because the bulls are longer than they are broad. It is plausible that the more difficult interpretation is intended. p. Amthor calculated in 1880 that the total number of cattle in this case is a number of 206. Also. the condition that the white and black bulls together form a square does not necessarily mean that their total is a square number.The Puzzles
9
The 'most complete' version contains the extra conditions that follow the ellipsis. then the solution is indeed complex and extraordinarily lengthy.

this has rather a lot. wishes to walk to the river for a drink and then back to T. Heron's assumption is correct. A modern version is the following: 18. Unlike the Chinese Tangram puzzle.10
Penguin Book of Curious and Interesting Puzzles
The Loculus consists of fourteen pieces making a square. (Is it a coincidence that the Tangram has the magical number of seven pieces and the Loculus exactly twice as many?) 17. To what point on the river bank should she walk?
. Assuming that light always travels by the shortest path. so it has important practical applications. T and C are mid-points. 75 AD). Mary. walking as short a distance as possible. and HK passes through A and OC through B. The object of the puzzle is to make figures with these pieces. where does it strike the mirror? This beautiful and important problem occurs in the Catoptrica of Heron of Alexandria (c. by bouncing off the surface of a plane mirror. How can this figure of an elephant be composed from the pieces of the Loculus?
Light Reflected off a Mirror
A ray of light passes from point A to point B. which might be said to have too few pieces. The method of division is almost self-evident: M. who is standing at S.

This is a great pity. Band C together in 20 days. How long will each require separately to do the same work?' Heron was also a geometrician:
21. 500 AD: 'I am a brazen lion. my spouts are my two eyes. My right eye fills a jar in two days. From The Tutorial Arithmetic by W. Find two rectangles.The Puzzles
11
-s
19. published in 1920: 'A and B together can do a piece of work in 6 days. my left eye in three. Workman. P. His Pneumatica describes scores of machines operated by wind and water. because the idea behind it is far from useless and turns up in many important situations. with integral sides. so it is no surprise that the famous cistern problem occurs in his Metrika.
. Famous? This is the infamous problem about the tank which is filled with water from several pipes. C and A together in 7t days. which was still being used to torture schoolchildren till the middle of this century. and which has become a byword for 'useless' mathematics. and the perimeter of the second is three times the perimeter of the first. and my foot in four. c. such that the area of the
first is three times the area of the second. From the Greek Anthology. tell me how long all four together will take to fill it?' Heron was a master of mechanical devices. 20. my mouth and the flat of my right foot. My mouth is capable of filling it in six hours.

the first known woman mathematician. and such problems in integers are named Diophantine after him. 'To find three numbers such that. What are the numbers? 25. the sum of the area and the perimeter is 280. Coincidentally. All his problems concern integers or rational numbers. 250 AD).12
Penguin Book of Curious and Interesting Puzzles
22. But when I first came upon the work of Diophantos. are 22.' The elementary problems that Diophantos solves could all have been presented. his method and his reasoning so overwhelmed me that I scarcely knew whether to think of my former self with pity or with laughter. 410). a commentary on his work was written by Hypatia (c. Find the sides and the area. and were by other writers. Xylander wrote in 1575: 'I came to believe that in Arithmetic and Logistic "I was somebody". but if the second receives 50 from the first. if each give to the next following a given fraction of itself. omitting each of the numbers in turn. to very difficult questions which had a stunning impact when his works were first translated into Latin and studied by European mathematicians more than 1200 years later. 8. the solutions are all whole numbers. The works of Diophantos vary from the simply puzzling and puzzlingly simple. the results after each has given and taken may be equal. Typically. What are the
. While studying such problems he is led to discuss the multiplication of positive and negative numbers. who was murdered by a Christian mob in the year 415. they are in the ratio 2:1. their ratio is then 1:3. as puzzles in everyday settings. in order. 'Let the first give! of itself to the second. 24. and the third give t of itself to the first. In a right-angled triangle with integral sides. had he so wished. I was adjudged an Arithmetician beyond the common order. 15 and 17 of Book I of the Arithmetica of Diophantos of Alexandria (c. respectively. among them some true scholars. What number must be added to 100 and to 20 (the same number to each) so that the sums are in the ratio 3:1? 24. The sums of four numbers. Two numbers are such that if the first receives 30 from the second. Here the numbers themselves are personified: 26. the second give i of itself to the third.
The First Pure Number Puzzles
23. 27 and 20. What are the numbers? These problems are nos. And in fact by not a few.

) Square Problems 27. many scientific principles can be expressed in terms of maxima and minima.. 'Given a long string. 29. and if we add 60 to this number. Find three numbers such that their sum is a square and the sum of any pair is a square. 30. 28. how should the string be disposed?' Here are two variants:
. some at 8 drachmas. II. the side of which is equal to the whole number of measures.' The Area Enclosed Against the Seashore So they reached the place where you will now behold mighty walls and the rising towers of the new town of Carthage. 'A man buys a certain number of measures of wine. Book I. with which to enclose the maximum possible area against a straight shore-line.. Find three numbers such that the product of any two added to the third gives a square. some at 5 drachmas each. Aeneid. 10). At a more serious level. and not in sequence. as Heron's problem of the ray of light reflecting off a mirror illustrates (see p. the result is a square. Find how many he bought at each price. Virgil. and they bought a plot of ground named Byrsa . 360-70 Questions and facts about extremes have a natural attraction. for they were to have as much as they could enclose with a bull's hide. He pays for them a square number of drachmas.The Puzzles
13
numbers?' (Diophantos assumes that all these transactions take place simultaneously. witness the runaway success of the Guinness Book of Records.

rho. composed of two identical halves hinged together. pyros. What is the maximum area of the triangle?
Metrodorus and the Greek Anthology The Greek Anthology is a collection of literary verses and epigrams. When will the area enclosed by the frame be a maximum?
32. Surprisingly. This figure shows the corner of a room with a screen. These include not only arithmetical puzzles. enigmas and puzzles. including beheadings. This frame is composed of four rods that ar'e hinged to each other at their ends. 500 AD). the son of Deidamia and the slayer of Polyxena. and this puzzle: If you put one hundred in the middle of a burning fire. but very early word puzzles. The answer is to put the Greek symbol for 100. Book XIV comprises a large number of riddles.
. hinged together.14
Penguin Book of Curious and Interesting Puzzles
31. How should the screen be placed to enclose as large an area as possible?
33. An isosceles triangle has two equal sides of length 10. into the word for fire. credited to Metrodorus (c. to get Pyrrhos. in which a word loses letter after letter from its front end but always remains a proper word. placed to cut off a portion of the corner of the room. you will find the son and a slayer of a virgin.

Later Arabic works tended to use the same expression al-jabr wa'l-muqabala. according to Islamic law. but let the fifth part of the legitimate one's share exceed by ten the fourth part of what falls to the illegitimate one. He clothed his cheeks with down. This is an essential study for lslamic
. '''Best of clocks. After consoling his grief by this science of numbers for four years he ended his life. Alas! late-born wretched child. Pyrrhus killed Priam. how much of the day is past?" There remains twice two-thirds of what is gone. Ah.The Puzzles
15
The arithmetical and logical puzzles include what were already classic problems. how great a marvel! the tomb tells scientifically the measure of his life. father of her second husband. and the cisterns problem. to refer to books on the same theme. chill Fate took him. or The Compendious Book on Calculations by Completion and Balancing. and Achilles killed her father. and adding a twelfth part to this. and my father-in-law my father. the day is counted as lasting for 12 hours. 'I desire my two sons to receive the thousand staters of which I am possessed. on the solution of equations. such as finding the weights of bowls given in arithmetical progression. after attaining the measure of half his father's life. from whence we eventually derive our word 'algebra'. He lit him the light of wedlock after a seventh part. or just al-jabr. Paris killed Achilles. al-Kitab al-mukhtasar hisab al-jabr wa'l-muqabala. and five years after his marriage He granted him a son. and new types: My father-in-law killed my husband and my husband killed my father-in-law. This tomb holds Diophantos.' (Problem 6.' (Problem 126) 36. Eetion. The answer is Andromache. 825 AD)
Al-Khwarizmi wrote a book. Achilles. 34. God granted him to be a boy for the sixth part of his life. my brother-in-law killed my father-in-law. Pyrrhus.' (Problem 11)
Arabic Puzzles
AI-Khwarizmi
(c.)
35. killed Hector. The second half of the same book deals with problems of inheritance.

Given three identical triangles. though as Ibn Khaldun wrote in the fourteenth century. meaning a compass which is so stiff that it can be used with only one opening. as it had previously been for Roman lawyers. How must the inheritance be divided? Abu) Wafa (940-998) Abul Wafa was born in Buzjan in Persia in 940. Dissect two identical larger squares plus one smaller square into one square. 1986. 'Some authors are inclined to exaggerate the mathematical side of the discipline and to pose problems requiring for their solution various branches of arithmetic. She also leaves t + j of her estate to a stranger.16
Penguin Book of Curious and Interesting Puzzles
jurists. leaving her husband. Dissect three equal squares into one
40. how can all four be dissected into one triangle? 43. How can two regular hexagons. He wrote commentaries on Euclid and Diophantos and AI-Khwarizmi. wouldn't they! This problem is practical: 37. The Rusty Compass Using only a straight-edge and a compass with a fixed opening. such as algebra. of different sizes. and similar things' (Berggren.53). larger. the use of roots. a son and three daughters. mathematicians would. regular hexagon? 42. According to law. and one smaller triangle similar to them in shape. 38. A woman dies. the husband receives one quarter of the ~tate and the son receives double the share of a daughter. 41. but he is best known for his study of geometrical dissections and of constructions with a rusty compass. Three Squares into One square. p. but this division is made only after the legacy to the stranger has been paid. construct at the endpoint A of a segment AB a
. be dissected into seven pieces which fit together to make one. so that one vertex is at a corner of the square and the other two vertices are on the opposite sides. Construct an equilateral triangle inside a square. 39. Well.

"I have asked for more wheat than you have in your entire kingdom. Oh King. who was asked by the Indian King Shirham what he desired as a reward for inventing the game of chess: 46. using only a straight-edge and a compass with a fixed opening ~qual to the radius of the circle. Using only a straight-edge and fixed-opening compasses. in this form: 47. women and children are there?
. let me cover each of the sixtyfour squares on the board.if it really dates as early as the third century .'" How many grains of wheat did Sissa require?
Indian Puzzles
The Bhakshali manuscript was found in 1881 in north-west India and dates from somewhere between the third and twelfth centuries. and four grains of wheat to place on the third. nay. for enough to cover the whole surface of the earth to the depth of the twentieth part of a cubit. '''Oh. Each man earns 3 coins. How many men. depending on which authority you choose. Construct a regular pentagon in a given circle. Twenty men. '''Majesty. 45. Sire. and so. you fool?" exclaimed the astonished King. without prolonging the segment beyond A.version of what came to be called 'One Hundred Fowls' problem (see Problem 74). give me a grain of wheat to place on the first square. 44. verily. divide a given line-segment into any given number of equal parts. each woman It coms and each child t coin. Sissa. and eight grains of wheat to place on the fourth." '''And is that all you wish. It contains the earliest . and two grains of wheat to place on the second square.The Puzzles
17
perpendicular to that segment.
Sissa and the Chessboard
Ibn Kallikan (c." Sissa replied. 1256) was the first author to tell the story of Sissa ben Dahir. women and children earn twenty coins between them. for more wheat than there is in the whole world.

. in a month of the spring season. the floor of a big mansion. the bitter. But instead of the total of 25 cents which they would have taken in operating separate enterprises.. Some (among them) broke down. Why?' 50. You tell me. 36 cents.. also formed a trust to sell their apples. and the amount of their pay was given to those that remained in the field. They had thirty apples apiece.'
51. one at two for 1 cent. in total. was lovingly happy along with her husband on . counting six times [in all] fell all of them everywhere.18
Penguin Book of Curious and Interesting Puzzles
Mahavira (c. Of this. each man obtained 10 puranas. Problems 48 to 54 are from his book the Ganita-Sara-Sangraha. 'Two market-women were selling apples. after thinking well. white like the moon. one-sixth fell on the bed.. cuckoos and bees which were all intoxicated with the honey obtained from the flowers therein. and resonant with the sweet sounds of parrots. at five for 2 cents.161 pearls. In order to end their competition they formed a trust. how many remained in the field and how many broke down. 'In how many ways can different numbers of flavours be used in combination together. and the other at three for 2 cents. then one-half of what remained (and one-half of what remained thereafter and again one-half of what remained thereafter) and so on. while under the old system they would have received a total of only 35 cents.' 49. 'Three puranas formed the pay of one man who is a mounted soldier. Then on a love-quarrel arising between the husband and the wife. who also had thirty apples apiece. and who were selling them at two for 1 cent and three for 1 cent. 850) wrote on elementary mathematics.. and if you know . 48. and at that rate there were sixty-five men in all. 'Two other women. give out the measure of the pearls. and there were found to remain [unscattered] 1. One-third of that necklace of pearls reached the maidservant there. their trust grossed only 24 cents. being selected from the astringent. together with the sweet taste?'
. a certain young lady .. pooling their stock and selling the apples at five for 3 cents. the pungent. the sour. 'One night. since under the new arrangement they took. This was to their advantage. and the saline. and situated in a pleasure-garden with trees bent down with the load of bunches of flowers and fruits. that lady's necklace made up of pearls became sundered and fell on the floor.

Arrows. as also the money on hand [with each of the three merchants]?'
53. circular in cross-section. "If I secure this purse." 'Then the third said. From the point where the two strings meet. How many [in all] are the arrows to be found [in the bundle] within the quiver?' 54. "I shall become three times as rich. one to the top of each.The Puzzles
19
52. Each of these two strings is stretched so as to touch the foot of the other pillar. Two pillars are of known height.) Bhaskara (IllS-c. What is the length of this suspended string?
This is identical to puzzles about ladders resting across passageways in which the heights of the points at which they touch are given. One said. the Lilavati. 131. can be packed in hexagonal bundles: 'The circumferential arrows are eighteen in number. "I shall become five times as rich. then the problem of finding the height of their intersection is far harder. from which the following problems
." 'What is the value of the money in the purse." 'Then the second said. Two strings are tied. (See p. 'Three merchants saw in the road a purse [containing money]. l18S) was an astronomer and mathematician whose most famous work. If not the vertical heights bur the lengths of the ladders are known. if they are thin cylinders. another string is suspended vertically till it touches the ground. I shall become twice as rich as both of you together.

the second has ten emeralds. and the bow: as those of Vishnu by the exchange of the mace. typical of the Indian style of the period: Joy and happiness is indeed ever increasing in this world for those who have Lilavati clasped to their throats. It ends with this delightful paragraph. Say. the rope. Each among them gives to each of the others two gems of the kind owned by himself. multiplication and involution. a king marched 2 yojanas the first day. and the third has eight diamonds. in a week?' 56. at a distance of thrice the pillar's height. the trident. the tabor. the dagger. the serpent. 'How many are the variations in the form of the God Siva by the exchange of his ten attributes held reciprocally in his several hands: namely. the elephant's hook. with what increasing rate of daily march did he proceed. What are the prices of those azure-blue gems. the arrow. since he reached his foe's city. Say quickly at how many cubits from the snake's hole do they meet. and then all three men come to be possessed of equal wealth. 55. or perhaps his wife. a distance of 80 yojanas. and lotus and the conch?' The final Hindu problem is unattributed. the discus. emeralds and diamonds?
. he pounces obliquely upon him. but on a popular theme: 58. pure and perfect as are the solutions. 'A snake's hole is at the foot of a pillar which is 15 cubits high and a peacock is perched on its summit. The first man has sixteen azure-blue gems. both proceeding an equal distance?'
It is natural that Hindu writers should have considered sooner or later the permutations and combinations of the attributes of their gods:
57. decorated as the members are with neat reduction of fractions.20
Penguin Book of Curious and Interesting Puzzles
are taken. Seeing a snake. intelligent calculator. was addressed to his daughter. gliding towards his hole. 'In an expedition to seize his enemy's elephants. the bedstead. the skull. and tasteful as is the speech which is exemplified.

required the number of each?' 61.
60. there would be a deficiency of 4 cash. well ahead of the West. who first controlled the flow of the Lo and the Yellow rivers. Required the respective numbers?'
. in Chinese legend going back at least to the fifth century BC.
The Nine Chapters
The Nine Chapters of Mathematical Art is supposed to have been written in the third century BC. 'A number of men bought a number of articles. it is only known that if each man paid 8 cash. and if each man paid 7 cash. was the gift of a turtle from the River Lo to the Emperor Yu the Great. which. 'Suppose that there are a number of rabbits and pheasants confined in a cage. How can the numbers 1 to 9 be arranged in the cells of this square so that the sums of every row and column and both diagonals are equal?
The resulting figure has essentially the arrangement of the Lo Shu. and contains the first known examples of the solution of linear simultaneous equations. there would be a surplus of 3 cash. as well as the extraction of square and cube roots.The Puzzles
21
Puzzles from China
The First Magic Square
59. neither of which are known. in all thirty-five heads and ninety-four feet.

when drawn towards the shore it reaches exactly to the brink of the pool. of which three bundles of the first class. the tip is just 3 feet from the stem. what is the length of the chain?' The following problem was also presented by the Indian mathematician and astronomer Brahmagupta. 'A chain suspended from an upright post has a length of 2 feet lying on the ground. one grows 3 feet and the other 1 foot on the first day.22
Penguin Book of Curious and Interesting Puzzles
62. The growth of the first becomes every day half of that
. as the Chinese called what we call Pythagoras's theorem. 'If five oxen and two sheep cost 10 taels of gold. and on being drawn out to its full length. what are the prices of the oxen and sheep respectively?' 63. so as just to touch the ground. And one of the first. its roots at the bottom of the pool. these are all set in remarkably realistic contexts. and two oxen and five sheep cost 8 taels. 'There are three classes of corn. 'There is a pool 10 feet square. three of the second and one of the third make 34 measures. 'Of two water weeds. the two short sides of which are respectively 8 and IS?' 68. 'There is a bamboo 10 feet high. what is the height of the break?' 67. Two of the first. with a reed growing vertically in the centre. two of the second class and one of the third make 39 measures. the end is found to be 8 feet from the post. which rises a foot above the surface. realistic that is if a mathematician happened to notice a reed breaking the surface of a pool. How many measures of grain are contained in one bundle of each class?' The following puzzles are from the ninth and last section of the book. the upper end of which being broken down on reaching the ground. two of the second and three of the third make 26 measures. more than 600 years later: 66. 64. In contrast to later problems in Diophantos. what is the depth of the water?' 65. or a chain hanging from a pillar. 'What is the largest circle that can be inscribed wlthm a rightangled triangle. and all concern right-angled triangles and the Gougu theorem.

when cycles of different lengths are compared.The Puzzles
23
of the preceding day. What will be the number?' 71. 'There are certam things whose number is unknown. Repeatedly divided by 3. The next problem is an example of the famous Chinese Remainder Theorem. by 5 the remainder is 3. in the Hai Tao Suan-Ching." . 'There are three sisters. the owner discovered the theft and pursued the thief for 145 miles. When he turned back the thief was riding 23 miles ahead of him. In how many days will all the three meet together?' Liu Hui (263 AD)." the woman replied. In how many days will the two grow to equal heights?'
Sun Tsu Suan-Ching (fourth century AD)
69. or The Arithmetical Classic of Ch-iu Chien (sixth century). When he had gone 37 miles. when an official whose business was overseeing the waters demanded of her: "Why are there so many dishes here?" '''Because a feasting was entertained in the house. the middle in every four days. rode away on its back. every three a dish for broth. Thereupon the official inquired the number of guests. who had stolen a horse. but every two used a dish for rice between them. Such problems had practical applications to calendar problems. the remainder is 2. If he had continued in his pursuit without coming back. What is the size of a square inscribed in the corner of a rightangled triangle to touch the hypotenuse? The Chang Sh'/u-Chien Suan-Ching." the woman said. of whom the eldest comes home once every five days. in how many further miles would he have overtaken him?'
. 'A man. while the other grows twice as much as on the day before. "how many guests there had been. poses one of the earliest chasing and returning puzzles: 73. '''I don't know. and by 7 the remainder is 2. and there were sixty-five dishes in all. he then returned. every four a dish for meat. poses this simple puzzle: 72. 70. 'A woman was washing dishes in a river. or Sea-Island Arithmetical Classic. and the youngest in every three days. [believing himself] unable to overtake him.

apparently. and this might be that collection. the cocks being sold for 5 shillings each. Hsu Ku Chai Chi Suan Fa (1275). the hens for 3 shillings and the chicks for :1 shilling each. Arrange the numbers 1 to 33 in these circles so that every circle and eve"ry diameter has the same total. among other matters. 1270 AD) wrote an 'Arithmetic in Nine Sections'. for pleasure'.24
Penguin Book of Curious and Interesting Puzzles
It also contains the earliest '100 fowls' problem:
74. 732-804). 'certain subtle figures of arithmetic.
'Propositions to Sharpen Up the Young'
Propositiones ad acuendos juvenes was written in the monastery of Augsberg. it is the earliest European collection of mathematical and
. It contains the following magic configuration: 75. and has been included in the works of Alcuin (c. the English scholar and churchman who spent his life at the court of the Emperor Charlemagne. which contains the very first extant representation of what we in the West call Pascal's Triangle (from an earlier Chinese source. His book was called. about the year 1000. Anyway. 100 fowls are sold for 100 shillings. 1000 AD). c. How many of each were sold? Yang Hui (c. on the grounds that Alcuin writes in one of his letters to the Emperor that he is sending him.

Say how many pigs must be killed each day. 'Two wholesalers with 100 shillings between them bought some pigs with the money. an odd number each day. in three days. 'An ox ploughs a field all day. But they couldn't. and selling them at the rate they had bought them for. of which ten were full of wine. to the second he went with one other . an odd number each day. They bought at the rate of five pigs for 2 shillings.' This is Problem XLIII. intending to fatten them up and sell them again. and contains the first appearance of many well-known puzzle types. they said to each other: "let's divide them". when dying. This puzzle is given to children to solve. and how could they be divided to make a profit.' This could be cruelty to little children. and the last ten were
. in such a way that from each manor he would take the same number of men as he had collected up to then. they made a profit. making a profit. Like the problems of Metrodorus in the Greek Anthology. By dividing them. How many pigs were there. The answer is: 'This is a fable.. When they saw this. which could not be made by selling them all at once?' (Problem VI) 78. But when they found that it was not the right time of year for fattening pigs. because they could only sell them for the price they had paid for them .. The servant went to the first manor alone. ten were half full. and orders that the pigs must be killed..' How many men were collected in all? (Problem XIII) 79. . they tried to sell them again to make a profit. a few riddles and trick questions find their way into the fifty-three problems in the collection. 'If two men each take the other's sister in marriage. 77. but it is also an early recognition that some problems simply cannot be solved. 76.The Puzzles
25
logical puzzles. gave to his sons thirty glass flasks. 'A kmg ordered his servant to collect an army from thirty manors. Nobody can solve how to kill 300 or 30 pigs in three days. How many footprints does he leave in the last furrow?' (Problem XIV) 'A man has 300 pigs. 'A father. what is the relationship between their sons?' (Problem XI) 80. and they were not able to feed them through the winter.

and the child's mother should receive one quarter. without sinking the boat. three. and so on every step up to the hundredth. and her mother fivetwelfths. How much should the mother receive. with two children who between them weigh as much as a loaded cart. each the weight of a loaded cart. she would receive seven-twelfths. appear for the first time in Alcuin. in which only two of them could cross at a time.' (Problem XIX) 84. each with a sister. How could this be done?' (Problem XVIII) 83. It happened however that twins were born . like the last two. if you can.a boy and a girl. On the first step stands a pigeon. But he had been ordered to transfer all of these to the other side in good condition. The only boat he could find could take only two of them at a time. five. 'A dying man left 960 shillings and a pregnant wife. a goat and a cabbage across the river. They find a boat which can only take one cartload. At the river. How did they cross the river. how much the son. How many pigeons are there altogether?' (Problem XLII)
. two pigeons. Make the transfer. and how much the daughter?' (Problem XXV) 85. Each one of them coveted the sister of another. four. needed to cross a river. He directed that if a boy was born. 81. on the fourth. But if a daughter was born.' (Problem XII) The next three classic problems. Three Friends and their Sisters 'Three men. a Goat. on the second. A Man. on the fifth. he should receive three-quarters of the whole. they found only a small boat. on the third.26
Penguin Book of Curious and Interesting Puzzles
empty. and a Wolf 'A man takes a wolf. so that each of the three sons receives equally of both glass and wine. have to cross a river. without any of the women being defiled by the men?' (Problem XVII) 82. A Very Heavy Man and Woman 'A man and a woman. 'A stairway consists of 100 steps. Divide the [wine] and the flasks.

He describes in the prologue to The Book of Squares how he was invited to the court of Emperor Frederick II of Sicily to compete in a mathematical tournament. 'Find a rational number such that 5 added to. or subtracted from. 116.produced from
. 'Three men possess a pile of money. which leads to the Fibonacci sequence: 88. so was often know.. 1.3688081075. He solved all three problems posed to him by John of Palermo. as Fibonacci (!ilio Bonacci). The first man returns 112 of what he took. the Liber Abaci. and learnt from the local Arabs the new Indian numerals. 'A certain man put a pair of rabbits in a place surrounded on all sides by a wall. How many pairs of rabbits can be .' The second was to solve the cubic
Xl
+ 2x2 +
lOx = 20
Leonardo found the solution. 1/3. The first. and how much did each man take from the pile?' Breeding Rabbits Fibonacci is best remembered for the following problem. which he helped to introduce to Europe. How much money was in the original pile. Practica geometriae. As a young man he travelled to Bugia in North Africa to help hi~ father. the second 1/3 and the third 116. 1175-1250) was a member of the Bonacci family. When the total so returned is divided equally among the men it is found that each then possesses what he is entitled to. who directed a trading post there.The Puzzles
27
Liber Abaci
Leonardo of Pisa (c. a compendium on geometry and trigonometry. is also a square. He wrote a book on calculations. in the style of Diophantos. which is correct to nine decimal places. Each man takes some money from the pile until nothing is left. was to: 86. its square. their shares being 1/2. This is the third problem: 87. our Hindu numerals. and the Liber quadratorum (The Book of Squares) on Diophantine problems.

giving each son one bezant more than the previous son and a seventh of what remained. 'A man entered an orchard [with] seven gates. but if you walk back too far. it will just appear small. How long does it take to climb out of the well? 93. A serpent lies at the bottom of a well whose depth is 30. how many hours would they take to devour it?' 90.) If you are too close. He continued in this way. The Best View of a Statue From what distance will a statue on a plinth subtend the largest angle? (See figure opposite. How many apples did he gather in the orchard?'
•
92. and there took a certain number of apples. a leopard would take five hours. then. 1370) discusses this famous puzzle. He did the same to each of the remaining five guards. It starts to climb. to his next son he left two bezants and a seventh of what was left. rising up 1 every day and falling back t at night.28
Penguin Book of Curious and Interesting Puzzles
that pair in a year if it is supposed that every month each pair begets a new pair which from the second month on becomes productive?' 89. Serpent Climbing out of a Well Dell'Abaco (c.
. and left the orchard with one apple. When he left the orchard he gave the first guard half the apples and one apple more. 'A man left to his oldest son one bezant and a seventh of what was left. we are asked. 'A lion would take four hours to eat one sheep. as a question about a suspended vertical rod. from the new remainder. How many sons were there and how large was the man's estate?' 91. if a single sheep were to be thrown to them. then. To the second guard he gave one half of his remaining apples and one apple more. the statue will appear greatly foreshortened. and a bear would take six. This problem was originally posed by Regiomontanus (1436-76) in 1471 to Christian Roder. to his third son he left three bezants and a seventh of what was left. from the remainder. It is notable as the first extremal problem since the days of antiquity and Heron's problem about the ray of light bouncing off a mirror. By this division it developed that the last son received all that was left and all the sons shared equally.

It is required to find in how many days they will meet. And from Rome to Venice is 250 miles. commanding him that he reach Venice in seven days. It happened that by order of these lords the couriers started on their journeys at the same time. a conversion of a try must be taken on a line extending backwards from the point of touchdown. most recently in this practical form: According to the rules of rugby union football. 'The Holy Father sent a courier from Rome to Venice. who should reach Rome in nine days. if the aim is to maximize the angle subtended by the goal-posts? This problem applies only when the try is not scored between the posts. From which point on this line should the conversion be taken.
The Couriers Meeting
The Treviso Arithmetic (1478) poses this problem: 94.The Puzzles
29
The same problem has been re-invented many times.'
The Nuns in their Cells
Pacioli in his De Viribus (c. at right-angles to the goal-line. And the most illustrious Signoria of Venice also sent another courier to Rome. 1500) posed this problem of rearrangements:
.

These two problems are from his Triparty en La science des nombres. How can you measure exactly 4 pints from a cask. Not wishing to abandon life. he proposed that they form a circle and that every third person. one in each cell. neither jar being marked in any way. making a total of three nuns along each side of the courtyard. Josephus. during the sack of the city of Jotapata by the Emperor Vespasian. published in 1484. 96. Liquid Pouring This problem first appears in Chuquet. given that you are allowed to pour liquid back into the cask? The Josephus Problem Chuquet was also the first to present an incident Josephus as a problem:
In
the life of
98. At the end of thirty days he finds he has paid out exactly as much as he received. There are eight nuns. You have two jars holding 5 and 3 pints respectively. counting round the circle. and also the best French mathematician of his time. How can they be rearranged so that there are four nuns along each side?
1 1 1
1
1 1
1
1
•
Nicolas Chuquet was a doctor by profession. How many days did he work? 97.30
Penguin Book of Curious and Interesting Puzzles
95. A carpenter agrees to work on the condition that he is paid £2 for every day that he works. should die. hid in a cellar with forty other Jews who were determined to commit suicide rather than fall into the hands of the Romans. in the order
. while he forfeits £3 every day that he does not work.

In other words. . So the second wife suggests that the children be placed in a circle and eliminated by counting continually around the circle. five. thirty children. and the second wife agrees. all her own children are then eliminated. By malice. six out . and how was the count done?
. . three out. One of the Christians suggests that all should stand in a circle and every ninth person counting round the circle should be chosen. are too numerous to share his inheritance.The Puzzles
31
in which they were selected. sons of the same father by his first and second marriages. in order to ensure that they were the last two remaining? When Roman troops were judged to have shown cowardice. are fifteen Christians and fifteen Turks. the historian of the Jewish struggle against the Romans. whence our expression 'to decimate' (which. The remaining child of that wife. tossed by storms and in danger of shipwreck. To lighten the load and save the ship. half are to be thrown overboard. confident that one of her children must be the last survivor. she ensures that the first fourteen children to be eliminated are all the sons of the first wife. the Turks: 99. once saved himself by just this trick. the count was: 'One. Later versions pitted Christians against their enemy of the period. The author describes how Josephus. two. seeing that he is alone. four. has come to possess the stronger meaning of 'reducing to one tenth' of the original number). suggests that the order of counting now be reversed. How were the children arranged. but to her mortification. In a later Japanese version. however. How should the Christians arrange themselves in the circle to ensure that only the Turks die? 100. On board a ship. and a companion who also wished to live. eliminating every nth child. based on an incident first described by the unknown author of the early work De Bello ludaico. they were lined up and every tenth man picked out for summary execution.' Where did he place himself. This snatch of military history may be the source for the Josephus problem.

102.32
Penguin Book of Curious and Interesting Puzzles
Exchanging the Knights
This is one of the earliest recreative chess problems. As a result the wine
. moving according to the rules of chess?
•
Niccoli> Fontana (c. A dishonest servant removes 3 pints of wine from a barrel. How can this be done? Tartaglia also gives a problem of this type: 104. These problems are from his General Trattato of 1556 and Quesiti et Inventioni Diverse of 1546. A man dies. posed by Guarini di Forli in 1512. and vice versa. He repeats his theft twice. leaving seventeen horses to be divided among his heirs. removing in total 9 pints and replacing them with water. How can the white knights take the places of the black knights. only to have Cardano wheedle the solution out of him and publish it himself. A man has three pheasants that he wishes to give to two fathers and two sons. was the brilliant mathematician who discovered how to solve the cubic equation. How can it be done? 103. giving each one pheasant. 1499-1557). Two white knights and two black knights are placed at the opposite corners of this portion of a chessboard.
~
~
~
~
101. nicknamed Tartaglia (the Stammerer). replacing them with water. in the proportions i: i:~.

and several think-of-a-number tricks. and tells you how many times 9 will divide into the answer. his Greek text of the Arithmetica of Diophantos (1621). who then gives the first three times as many counters as the first has in his hand. You can now say that the second person has . It contains river-crossing problems originating with Alcuin. he adds one and then takes one half. and the first European work devoted to mathematical recreations. 4 and 5. Weights for use with a balance were traditionally made in nested form. He is famous for two works. which are presented here in the form of problems: how is the original number recovered after the following operations? 105. he takes half of it. or if it is odd. so that one weight fitted inside the other and the largest weight contained all the smaller weights.what? 107. separately.how many . the Josephus problem as solved by Tartaglia.what? 106. and one of the earliest members of the Academie Fran\. Problemes plaisans et delectables qui se font par les nombres (1612). accompanied by his own Latin commentary. The number he chose is . a liquidpouring problem. The Problemes plaisans was largely a compendium of previous puzzles. a method of constructing magic squares which is that found in Moschopoulos. and the second person takes three times as many. greater than 5. and trebles it. The subject chooses a number less than 60 and tells you the remainders when it is divided by 3. ignoring any remainder. The original number is . A person chooses secretly a number. not successively. telling you whether the product is odd or even.aise as well as a mathematician. Next he multiplies the result by 3. The first person secretly chooses a number of counters.counters in his hand?
Problemes plaisans also contains the famous problem of the weights. The first then gives 5 counters to the second. If it is even. How much wine did the barrel originally hold?
Bachet
Claude Gaspar Bachet de Meziriac (1581-1638) was a poet and translator. Modern sets of weights in which each fits snugly into the top of the next in the series are a variation of
.The Puzzles
33
remaining in the barrel is of half its former strength.

though it is widely suspected that it is true. were really necessary to weigh a given quantity. In his commentary on Diophantos VI.a)(s .' This theorem became known as Fermat's Last Theorem. if the weights can be placed in either of the scale pans? Bachet's Diophantos is most famous because it was in the margin of his own copy that Pierre de Fermat wrote a comment on Diophantos's Book II. one of whose altitudes and its three sides are consecutive numbers? 110. which means that the triangle also has a rational area. which are not rightangled. Bachet asked: 108. problem 8. The area of a Heronian triangle is always a multiple of 6. I have discovered a truly remarkable proof which this margin is too small to contain. Or a Collection of sundrie excellent Problemes out of ancient and
. What are the three Heronian triangles. What is the unique Heronian triangle with area 24?
Henry van Etten
Henry van Etten (1624) was the author of Mathematical Recreations. 18. Since all the measurements can be multiplied up to make them integers. Heronian triangles are often considered to have integral sides and area.
A = Js(s . It was natural to wonder how many weights. and remains unresolved to this day. or in general any power higher than the second into two powers of like degree. Because the area of a triangle can be calculated from the sides using Heron's formula. What are the sides and area of the unique Heronian triangle.c)
where s = !(a + b + c). What is the least number of weights that can be used on a scale pan to weigh any integral number of pounds from 1 to 40 inclusive. 109.b)(s . to solve xr + yZ = a Z in integers: 'it is impossible to separate a cube into two cubes. Bachet asked for a triangle with rational sides and a rational altitude.34
Penguin Book of Curious and Interesting Puzzles
this. or a biquadrate [fourth powers) into two biquadrates. and which weights. whose area and perimeter are equal? 111. such triangles are called 'Heronian'.

Arrange three knives so that they 'hang in the air without being supported by anything but themselves'. 113. Its mathematical problems are mixed up with mechanical puzzles and experiments in optics and hydrostatics. including questions from the Greek Anthology and copying from Bachet. The solution is to hit it sufficiently sharply in the middle. on which it was certainly based.The Puzzles
35
moderne Phylosophers Both usefull and Recreative. You have a strong staff. How maya man have his head upwards and his feet upwards at the same time?
. but containing much extra and varied material. and a bucket almost full of water. It was a compilation. 115. in the ample space between three other drinking glasses placed on the table with more than enough space for a fourth glass to be placed on the table between them. Variants 111 Victorian puzzle books demanded how three knives might be used to support a drinking glass. The first mechanical problem is to break a staff resting on two glasses of water. Several of the following problems also appear two centuries later as popular Victorian amusements. one square. and tips such as 'How to keep wine fresh without ice or snow in the height of summer'. How can a bottle be lifted using only a single straw? 116. Required to support the bucket over the edge of the table. due to the inertia of the staff.
112. and it will break. attributed to Aristotle. instructions on the making of fireworks. What shape of bung can be used to plug three different holes. published in French in 1624 and first published in English translation in 1633. one triangular and one circular? 117. How can a stick be made to balance securely on the tip of a finger? 114. It is also an important early work on conjuring. which allows a variety of drinks to be poured at the magician's whim from the same spout. containing the first description of the 'Inexhaustible Barrel' or 'Any Drink Called For'. naturally.

How was this possible? 123. each having sold all their apples. and C. Where can a man look south in all directions? 120. Find the point whose sum of distances from the vertices of a given triangle is a minimum. Explain. because it can be interpreted as asking for the shortest road network that will join three towns at the vertices of the triangle. How could this be?' 124. A. Two men ascend two ladders. who invented the barometer: 125. but did not live to the same age. 'Three women. A strip of paper can be transformed into a pentagon. B. a soldier in the Civil War and an inventor and early member of the Royal Society. He posed the following problem to Torricelli. Galileo's famous pupil. and then died at the same time. The next problem occurs first in Urbino d' Aviso's treatise on the sphere (1682): 126. How can an oval be drawn with one turn of the compass? 122. brought home as much money as each other. and who leapt beyond Diophantos to found the modern theory of numbers. B. travelled the world. He enquired:
. A had 20. at the same speed. 119. C. and yet they get further apart. This problem has a natural appeal. carried apples to a market to sell. Two horses were born at the same time. and.36
Penguin Book of Curious and Interesting Puzzles
118. the one as the other. How? Prince Rupert's Cube Prince Rupert was a nephew of Charles I of England. Why must there certainly be at least two people in the world with exactly the same number of hairs on their head?
•
Pierre de Fermat (1601--65) was a lawyer by profession and an amateur mathematician of genius who contributed to the development of the calculus and the invention of analytical geometry. How can a compass with a fixed opening be used to draw circles of different sizes? 121. 30. they sold at the same price. 40.

What IS the largest cube that can be passed through a square hole cut in a given cube? Sir Isaac Newton (1642=1727) composed a book on elementary algebra.The Puzzles
37
127. or at least 3 sixes with 18 dice? The misaddressed letters Niclaus Bernoulli (1687-1759). his Arithmetica Universalis (1707). 132. In how many ways can all the letters be placed in the wrong envelopes? 131.
what is the relationship between the nine magnitudes a to c"? In 1693 Samuel Pepys the diarist and Secretary for the Navy wrote to Newton with this query. If a cows graze b fields bare in c days. in which this problem occurs: 128. on average. a natural question for a gambler: 129. Here are three of the problems he considered. A related question: if seven letters are placed in seven envelopes randomly. and a' cows graze b' fields bare in c' days. to find in their correct envelopes?
•
Leonhard Euler (1707-83) was one of the most versatile mathematicians of all time. to throw at least 1 six with 6 dice. considers this problem (but with n letters instead of ten): 130. one for each letter. visiting each square once and only once.so that the circuit is continuous?
. how many letters would you expect. and ending up a knight's move from its starting square . one of the extraordinary Bernoulli family which produced nine outstanding mathematicians in three generations. as well as one of the greatest. 38. and a" cows graze b" fields bare in c" days. A correspondent writes ten letters and addresses ten envelopes. The Knight's Tour How can a knight make a complete tour of the chessboard shown on p. Which is more likely. at least 2 sixes with 12 dice.

A vertex where an odd number of edges meet is called an odd vertex. Graph theory poses many problems. joined by a set of lines. crossing each bridge once. called vertices or nodes. bur not crossing any bndge twice? This was the first ever problem in what is now called graph theory. A graph is a set of points.38
Penguin Book of Curious and Interesting Puzzles
The Bridges of Konigsberg In the town of Konigsberg there were seven bridges across the river Pregel. some of them very simple and simply puzzling:
. Is it possible to go for a walk. naturally. called edges. This popular question was answered by Euler in 1736:
133.

How can five each of As. but the same problem . comprismg a colonel.with twenty-five officers IS not: 135. This problem turns out to be impossible. lieutenant and sub-lieutenant from each of six regiments. Ds and Es be placed in these cells so that no letter is repeated in any row or column?
The Ladies' Diary or Woman's Almanac. major.
. sketches of notable women. to be replaced by rebuses. Why is the number of odd vertices in a graph always even?
39
The Thirty-six Officers Problem Euler considered the problem of placmg thirty-six officers. Cs. and proof that caricatures of women as mathematically incapable were less well-established in the early eighteenth century than the late twentieth century. captain. enigmas and mathematical questions. That it not only survived but flourished is a blow in the eye to those who suppose that women cannot be interested in mathematics. lieutenant-colonel.The Puzzles
134. however. Bs. Although men soon proposed and answered many of the questions. its contents changed. naturally appealing to its readership. in a square array so that no rank or regiment will be repeated in any row or column. 1704-1841
The Ladies' Diary was first published in 1704 and consisted initially of recipes. Within a short time. and articles on education and health.

it is a variant of problem 66 and more
. such as 'Anne Philomathes') concerned grains of wheat on a chessboard. a professor at the Royal Military College. from its Commencement in the Year 1704 to 1816 was a compilation. 'From a given cone to cut the greatest cylinder possible. in the manner of the times. The problems were initially proposed. One half. one third. many of the questions were very difficult. some venerable problems were proposed. and their Original Answers. as payment for sixty-four diamonds: one grain on the first square. The Mathematical Questions Proposed in the Ladies' DIary. 'In how long a time would a million of millions of money be in counting. The very first mathematical question. and the year to consist of 365 days.
If to my age there added be. was: 136. this practice was soon abandoned. Pray find out what my age may be. posed in the year 1707. the tip making a mark on the ground. 5 hours. together with Some New Solutions. Six score and ten the sum you'll see. and three times three. but. The first question that is identified as 'Solution by a Lady' (respondents were often anonymous or identified by aliases. In other words. the work of Thomas Leybourn. In subsequent years. 'A person remarked that upon his wedding day the proportion of his own age to that of his bride was as 3 to 1. two on the second and so on. 45 minutes?' This early question in verse illustrates the difficulties of the form: 137.' Question 42 concerns a maypole which breaks. What were their ages upon the day of their marriage?' Question 36 was posed by Mrs Barbara Sidway: 139. in verse.40
Penguin Book of Curious and Interesting Puzzles
women continued to contribute as posers and solvers. mathematics not lending itself to versification. and were answered by almost all the famous English mathematicians of the eighteenth century. Although it was not a compilation. supposing one hundred pounds to be counted every minute without intermission. but fifteen years afterwards the proportion of their ages was 2 to 1.
138.

proposed the first of many vanishing square paradoxes. What is the least number which will divide by the nine digits without leaving a remainder? 141. in his Rational Recreations (1774). In contrast the next three problems have a modern feel: 140. they brought their wives with them. likewise.
8 5
5
3 3
8
8
. What was the time?
•
The Vanishing Square Paradox
William Hooper. It was evening. Catriin. the women's Geertrick. that I could not distinguish the figures. 'They told me that they had been at market. The men's names were Hendrick. 'Being at so large a distance from the dial-plate of a great clock. Question 51 was also old. and Cornelius.' the correspondent noted that they were in a straight line and pointing upwards to the right.) 142. and Anna. each person bought as many hogs as they gave shillings for each hog. 'There came three Dutchmen of my acquaintance to see me. to buy hogs. and Claas bought eleven more than Geertrick. each man laid out 3 guineas more than his wife. I desire to know the name of each man's wife?' (A guinea was 21 shillings.The Puzzles
41
than 1000 years old. but I forget the name of each man's wife. Book V. The solution noted that the problem appeared in Diophantos. Claas. Hendrick bought twenty-three hogs more than Catriin. but as the hour and minute hands were very bright and glaring. being lately married.

The top square has an area 8' = 64. the boats are equally laden. Your task is to reassemble these sixteen pieces to make another 13 x 13 square. when. but with an empty square in the centre. form a rectangle 5 x 13 = 65. 145. the rowers equally good. at the same time. and.42
Penguin Book of Curious and Interesting Puzzles
143. another should set out from Newburyport. during ebb tide. suppose the current forwards one and retards the other 1t miles per hour. to come down the river. but not collected
.
Rowing with and against the Tide
This is another first. to go up the river. and. allowing the difference to be 18 miles. Where has the extra square come from? 144. will the two boats meet?'
Rational Amusements for Winter Evenings
John Jackson was 'A Private Teacher of Mathematics' who decided that there were many puzzles scattered around. This is a modern variant. when reassembled to make the lower figure. The same four pieces. 'If. in the common way of working in still water. a wherry should set out from Haverhill. would proceed at the rate of 4 miles per hour. which occurs in an arithmetic textbook published in the United States in 1788. in the river.

4 .' 151. there then. Geography.' (In other words. or. The true result (it has been tried) Exactly is eleven..)
. which is a shame because.The Puzzles
43
together in one small and convenient volume.. How can you succeed? 152. &. a right-angled triangle. to be assembled to form a square. By just one fifth of seven. A Collection of above 200 Curious and Interesting Puzzles and Paradoxes relating to Arithmetic. And if fifty from forty be taken. and the product shall still consist of nine different (digits]. a rectangle. Geometry.' 147. Shall just half a dozen remain. 146. a rhombus. which appeared in London in 1821.'
148. 'It is required to express tOO by four 9s. the products of the numbers of each vertical column and each horizontal row in the box overleaf must equal 4096. in addition to its superb title anticipating the present volume! . 'Designed Chiefly for Young Persons'. 'One third of twelve if you divide. 'Place in a row nine (digits] each different from the others. and from nine ye take ten (Ye youths. so he assembled them himself and wrote Rational Amusements for Winter Evenings. You have 12 pints of wine in a barrel and you wish to divide it into 6 pints for a friend and 6 pints for yourself. to form a product of 4096 each way. now the mystery explain).
'If from six ye take nine.'
150. 'Place the nine digits. From its great'rarity it may be inferred that it did not sell many copies. and so on. so that the sum of the odd digits may be equal to the sum of the even ones. 2.' 149. but you only have containers holding 7 and 5 pints. 'With the first nine terms of the geometrical progression 1. in particular a collection of ten tree-planting problems and a collection of fifteen variously shaped tiles which resemble a complicated set of Tangram pieces.it contains many of the classic puzzles and some that had not apparently appeared earlier in print. Multiply them by 8.

the name of which is now required. 'Mathematicians affirm that of all bodies contained under the same superficies. a sphere is the most capacious.3. greatest breadth 4 inches. and greatest depth 3 inches.'
154.. of which it may be truly affirmed that supposing its greatest length 9 inches.2. but they have never considered the amazing capaciousness of a body. to form 34 every way. 'A Cheshire cheese being put into one of the scales of a false balance.44
Penguin Book of Curious and Interesting Puzzles
153. to 16. yet under these dimensions it contains a solid foot?'
.. . . was found to weigh 16 lbs. and when put into the other only 9 lbs. What is the true weight?' 155. 'With the numbers 1.

as well as in longitude. where. the Isle of Dogs over against it. when the tide is in. 'There is a certain village in the Kingdom of Naples. situated in a very low valley. and yet. every noon by 3000 miles and upwards. 'There are three remarkable places on the globe that differ in latitude.' The next question asks the reader to inscribe a regular dodecagon (12 sides) in a circle under the same conditions. than when he either rises or sets. not one can be seen. and is worth relating to illustrate how puzzles in natural philosophy were often mixed in with more logical or mathematical puzzles: 'There is a certain place in Great Britain. when the tide is out.' 160. that island is farther from France than England is. 'To make a triangle that shall have three right-angles. where the winds (although frequently veering round the compass) always blow from the north point.' 162. 'There is a particular place on earth. 'There is a certain island.The Puzzles
45
156.' jackson's explanation is: 'The place may be the wharf at Greenwich. certainly well-calculated to enliven lessons on the globe. situated between England and France.' The remaining puzzles require no such localized knowledge. the sheep may be seen feeding on a certain neighbouring island. One of these depends on acquaintance with a specific physical phenomenon. 'To inscribe a square in a given circle.' 161. all of them lie under the same meridian. and the water at the lowest. supposing the centre to be known. by means of compasses only.' 158. yet. and yet. and yet the sun is nearer to the inhabitants thereof. and the appearance caused by refraction. though they be feeding there at the same instant.'
.' 157. 'Divide a circle into four equal parts by three lines of equal length. when the water is high. to those of the said village. Jackson'S book concludes with no less than sixty 'Geographical Paradoxes'. 159.

The Turks the sixth.... Have each his own true sabbath. [Such tiles) form the object of a pastime. The following problems are taken partly from Ozanam's 1741 edition. I pray. The hour and minute hands of a watch coincide at noon. How is this possible?
From Ozanam to Hutton
Jacques Ozanam was a Frenchman. called by the French Jeu de Parquet . he was induced to try in how many different ways they could be joined side by side. having a border round it. How can these three.
164. professor at the Royal Military Academy at Woolwich. and day. a small table..46
163. A traveller sets out on a journey.. some square porcelain tiles. The Jews the seventh. and yet his head remains attached to his body. During his journey. in a memoir printed [in) 1704 . When will they once again coincide. and capable of receiving sixty-four or a hundred small squares . during the next 12 hours? 166. It was greatly enlarged and improved by Jean Etienne Montucla (born 1725).. tell. divided by a diagonal into two triangles of different colours . as we have oft been told. of the Royal Academy of Sciences.. in the same place. 165. and eventually returns to the place from where he started. called Recreations in Mathematics and Natural Philosophy (1840). 'We are told by Father Sebastian Truchet. Edward Riddle's revision of it. born in 1640.. with which people amuse themselves in endeavouring to form agreeable combinations. was the largest collection of mathematical puzzles and recreations published in this country up to that time. How many figures can be formed by three squares if the colours of the two halves are black and white and if an edge is placed against a complete edge?
. who wrote a book of recreations based on Bachet and other traditional sources. his head has travelled 12 yards further than his feet. It was translated into English by Charles Hutton. that having seen during the course of a tour which he made to the town of Orleans. as they did of old. in order to form different figures. and partly from Riddle.
Penguin Book of Curious and Interesting Puzzles
Christians the week's first day for sabbath hold. a friend of Diderot and D' Alembert.

who said that he got it from M. If A and B together can complete a task of work in 8 days. DIssect a gIven rectangle into a square. requests to know the proper dimensions of the vessel. one of which sells at 10 shillings per bottle. but being desirous to save the material as much as possible. capable of contaIning a cubic foot of liquor. the pieces to be moved by transposition only. 'A gentleman wishes to have a silver vessel of a cylindrical form. 171.The Puzzles
47
167. How can five equal squares be dissected and reassembled to form one large square? 169. and Band C together take 10 days. 170. and the other at 5 shillings. A man has two wines. open at the top.' 172. Demonstrate Pythagoras's Theorem by dissecting the smaller squares to form the larger square. without rotation or turning over. and if A and C together take 9 days. What is the mixture that would sell at 8 shillings a bottle? 173. how much will each man take to do the work by himself? The next problem is equivalent to the dissection of a Greek Cross into a square: 168. the French naturalist and translator of Newton's
PrincIpia:
. BuHon. What is the largest rectangle that can be cut in one piece from this triangular piece of timber?
The next problem is attributed to a Mr D-.

the arithmetic and the geometric. In other words the harmonic mean of x and y is
1
which simplifies to
2xy x+y
With that explanation. The harmonic mean of two numbers is found by taking their reciprocals. 'Given two lines and a point within the angle formed by them.' 176. The
. to find the smallest triangle by area that can be cut off. which are well known. 177. and the harmonic. The Harmonic Square The ancient Greeks considered three important means.48
Penguin Book of Curious and Interesting Puzzles
174. One player chooses a number less than 11. and this process is then repeated. in four different ways. how can the cells of this square be filled so that the cells in the middle of each side and the centre cell are each the harmonic means of the numbers sandwiching them? The central number is sandwiched. the mid-points of the sides are joined in sequence. The second player does likewise and adds his number to the first player's number. Given any irregular polygon. finding their average and taking the reciprocal of the result.' 175. of course. 'It is required to find the point where these divisions will terminate. again and again.

in 1848. Enigmas. Is there a winning strategy? 178. They were mostly written for young people. in the chess magazine Berliner Schachzeitung. with an extra condition:
. Charades. Card Games and Fireside Fun. simple scientific experiments. published in 1821'. enigmas. Parlour Magic etc. wntlng under the pseudonym 'Schachfreund'.ational Amusements: 180. The Eight Queens How can eight queens be placed on a chessboard so that no queen attacks any other? This problem was first posed by Max Bezze!. FireSide Games. Cassell's Book of Indoor Amusements. because of the size of the board. plus a wealth of literary puzzles. and contained sections of mathematical. and the end of the century.The Puzzles
49
first player again adds a number less than 11. and The Illustrated Book of Puzzles and Parlour Pastimes: A Repertoire of Acting Charades. According to Dudeney. often a section of magic tricks. which was John Jackson's P. Riddles. writing in his The World's Best Puzzles. but Dudeney himself knew of no earlier source than 'a rare book.6) queens be placed on a 4 x 4 (5 x 5. rebuses and maybe chapters on outdoor as well as 111door games and amusements. and so on. Tree in a Row How can nine trees be arranged in ten rows with three trees in every row?
Parlour Pastimes posed a variant in verse. charades. The player who reaches the grand total of 100 or more is the winner. The authors showed the usual reliance on old sources. 6 x 6) board so that no queen attacks any other?
The Victorian Era
Between John Jackson. Conundrums. a wealth of books appeared. How can 4 (5. An easier problem is: 179. which IS why the problems below are not credited to particular books. with titles such as The Games Book for Boys and Girls. Arithmetical and Mechanical Puzzles. Riddle's edition of Ozanam. this puzzle has been attributed to Sir Isaac Newton. To find all the solutions is extremely difficult. mechanical and word puzzles.

as under. passing over two each time. can thiS be so?
. the product of the multiplication. is a simpler problem: 182. on the face of it. and the fourth part you divide by 2.' 184. pray dispose.) 187.50
181. 'Draw six lines as under.' 2 3 4 5 678
186. A plan of it I fain would have. as I can show. the remainder of the subtraction. show how they can be laid or placed in four couples. 'Place twelve counters in six rows in such a manner that there shall be four counters in each row. add five other lines. 'Having placed eight coins in a row. Three trees I'd have in every row.' 183. the third part you multiply by 2. You have to divide the number 45 into four parts. so that the sum of the addition. A pond in the midst I'd have also. Twenty-four trees in twenty-eight rows.
The following. The half of twelve IS seven. Which makes me your assistance crave. and the quotient of the division are all equally and precisely the same. How is this possible? 185. 'Plant four trees at equal distance from each other. The half of thirteen eight.
Penguin Book of Curious and Interesting Puzzles
Ingenious artist. To the first part you add 2. from the second part you take 2. and make the whole form nine. removing only one at a time.'
(There is also a French version of this puzzle: add three lines to make eight.

The Puzzles
188. or paper. form a cross.'
. cut in the following shapes. Add four lines. These dogs are dead.
51
189. 'Of five pieces of wood. and then they'll run away. perhaps you'll say.

How can an equilateral triangle be constructed by folding a square of paper? 193. and it was necessary so to divide it. 'A charitable individual built a house in one corner of a square plot of ground. From these five shapes. 192. In the ground were four cherry trees. How can a ladder be made out of a single sheet of paper. based on the popular French book La Science Amusante.52
Penguin Book of Curious and Interesting Puzzles
190. and let it to four persons. Here is a sketch of the plot. that each person might have a tree and an equal portion of garden ground. and this to be effected with three cuts with the scissors?
. without lIsing gum or other adherent. also form a cross:
191. How is it to be divided?'
The next four problems are from Scientific Amusements by 'Tom Tit'.

. that you pass over just two coins.1 cm across and the old shilling was approximately 2. Repeat this till there is no single coin left. Cut a hole in a visiting card large enough for a person to climb through.' How can he achieve this feat. Arrange the digits 0 to 9 so that they sum to 100. be cut into four identical pieces. marked as shown. place it upon some other. Then. and no mark is cut?
o o
o
00 00
o o o
o
o
198.• A carpenter had to mend a hole in the floor which was two feet wide and twelve feet long. 195. The board given him to mend it was three feet wide and eight feet long. cutting the board into only two pieces? 197. How can a half-crown be passed through a hole the size of a shilling? The old half-crown was approximately 3. How can this board. with this proviso. taking up anyone of the series.The Puzzles
53
194.3 cm in diameter. 200. Place ten coins in a row upon a table. so that each piece contains three of the marks. 199. What is the largest envelope that can be constructed by folding a rectangle of paper? 196.

How many animals are concealed in this picture?
202. without going over any line twice? A stroke ends as soon as you lift your pencil from the paper. How many strokes are necessary to draw this figure. so that each receives an area identical in shape and size to the others'? 203.
. This is another square with one quarter missing.
How can it be divided between four sons.54
Penguin Book of Curious and Interesting Puzzles
201.

but you are not allowed to use a knife. and moving only three others. and the opposite corner in the other hand. Under the centre of the tumbler lies a 20p coin. How do you remove the 20p piece? 205. you bring the corners together. Here are the same three squares of matches.
206. A tumbler is resting on three lOp pieces on a table cloth. Taking one corner of a plain unknotted handkerchief in one hand.
. and then apart.
Move three matches to show what matches are made of. Here are three squares. and Lo and Behold! there is a knot in the handkerchief! At no time did you release either of the corners of the handkerchief. each composed of four matches. Being thinner than the lOp piece.The Puzzles
55
204. or sheet of paper or card or any other suitably thin instrument.
DDD DDD
207. it can in theory be removed from under the edge of the tumbler without disturbing the tumbler. Make them into one by taking one match away.

except one. which shall be the 9.IS
lOY-
12'10
209. An Easy Solitaire Each number represents a piece that can jump over any other piece.
3/8
303
300 2. horizontally or diagonally. How can all the pieces be removed. Here is a correct addition sum. How is this possible? 208.. exactly as you picked it up. which ends up in its original position in the centre?
1
8
7 6
2
9 4
3
5
. either vertically. to leave a new addition sum which still totals 1240. Your puzzle is to cover one of the numbers completely. into an empty square beyond.56
Penguin Book of Curious and Interesting Puzzles
which remain between the fingers of each hand.

that which you Are striving to unfold An octagon. I should have said of each shape four. Remove just three counters to leave just three equal squares. familiar to My friends both young and old.
0 0 0
0 0 0
0 0 0
Upon a piece of paper draw The three designs below. How can four triangles be made with just six matches?
57
211. If joined correctly.The Puzzles
210.
3
.
0 0 0
212. These twelve counters are arranged to form six equal squares. Which when cut out will show.

which crammed 794 puzzles. father says. Bet. Who Can Tell? Twice six are eight of us. And every week beats jim by pennies five. And in this way: The eldest daughter. selected by Don Lemon (1890). as was the delight in quibbles and trick questions. Earn ten and tenpence. There was a poor man called johannes Bull. Some tell-tale facts I'll now disclose to you. into 125 pages and included most of the puzzles above. jane. Who children did possess. Six little workers had he. Gains seven pence more than sister Anne can gain. I will not tell you how much each did gain. now do you see?
. Bet. each of whom Earned something for the household at the loom. Week after week. a quiver full. jane. The expression of puzzles in verse was typically Victorian. Rose. And who yet managed somehow to scratch on. Now. a large majority of them word puzzles. though not up to jane. Rose. But. Twelve are but six of us.58
Penguin Book of Curious and Interesting Puzzles
The most comprehensive compilation of this period was Everybody's Illustrated Book of Puzzles. while joe can get By his endeavours sixpence more than Bet. But by her hands gets fourpence more than Rose. Ann. for him. as I would you should possess a clew. earns not so much as those. What can we possibly be? Would you know more of us? I'll tell you more of us. By the true help of daughter and of son. and I don't speak plain. yet means to thrive. not so old. joe. Six are but three of us. and the following. Nine are but four of us. jim. For I'm a puzzler. 213. say what each child worker should receive When father draws the cash on pay day eve? 214. Ann eightpence more than joe. Five are but four of us.

having a coin in each hand. you'll find very fair The sum will be nothing. and without bringing my hands together.
59
216. A box has nine ears of corn in it. in truth I declare. A row of four figures in value will be Above seven thousand nine hundred and three. what will the fourth of twenty be? 217. Quibbles
(a) Add the figure 2 to 191 and make the answer less than 20. How is this explained? 218. having a garden attached to the house. he wished to divide it among them. though his legs should be free? (d) If five times four are thirty-three. But when they are halved. How did he do it?'
! !
.The Puzzles
215. (b) How can I stretch my hands apart. and yet it takes him nine days to carry the corn out. A squirrel carries out three ears a day. cause both coins to come into the same hand? (c) How must I draw a circle round a person placed in the centre of a room so that he will not be able to jump out of it. 'A person let his house to several inmates and. There were ten trees in the garden and he desired to divide it so that each of the five inmates should have an equal share of the garden and trees.

6. How can this rectangle with two tabs be cut into two pieces to make a complete rectangle?
221. price one guinea. and walking off with the hat. The hatter gets the note cashed by a neighbour. Query: what will the answer tell the younger person?
The Learned Professor Hoffman
Professor Hoffman's real name was the Reverend Angelo John Lewis. to find a number of six digits of such a nature that if you transfer the two left-hand digits to the opposite end. whose sum shall be equal to unity. 1. His most famous book. then ask the older person to add this difference to their own age. 9. This is a trick for a teenager addressing an adult. and then to take the first digit of the amount and add it to the remaining figure. 2. 'Required.' 222. He also wrote on magic and conjuring. the purchaser pocketing his change. 4. than the neighbour comes in with the news that it is counterfeit.60
Penguin Book of Curious and Interesting Puzzles
219. the new number thus formed is exactly double the original number. 'A man goes into a shop and buys a hat. He offers in payment for it a £5 note. No sooner had he left. was chiefly devoted to the many popular mechanical puzzles but he also included other Victorian favourites. 7. 3. and once only. to compose two fractions. Let the teenager subtract his or her age from 99. 'Required. 0. £3 19s. 5.'
. Puzzles Old and New (1893). and the hatter has to refund the value. however. 220. of the numbers. Each number to be used once. 8.' 223.

'
225.' It afterwards transpired that he had dined alone. 'there was my father's brother-in-law. How could this be?
LV\L\
~
/\
. and my brother-m-Iaw's father-in-law. my father-in-Iaw's brother-in-law. An old gentleman was asked who dined with him on Christmas day. we were qUite a family party. 'Fifteen matches being laid on the table so as to form five equal squares.' he replied. How can three matches be taken away to leave a total of seven triangles behind?
226. 'Well.) 224.The Puzzles
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How much is the hatter out of pocket by the transaction? (A guinea was 21 shillings. and yet his statement was correct. to remove three matches so as to leave three such squares only. my brother's father-in-law. required.

Each had a herring. 229.
Hamilton and the Icosian Game
The Icosian Game was invented by W. Five herrings were divided between five persons. How was this possible? Mathematicians have often created problems of a popular nature in between their more 'serious' work. with a few letters omitted. Euler was an example. How can this be done? A solid model is not necessary: the arrangement of the towns is indicated 111 this figure. for £25. As 'The Traveller's Dodecahedron' it consisted of a regular dodecahedron. in alphabetical order from B for Bruxelles to Z for Zanzibar. handsomely made in wood. Jacques and Son. What is the difference between six dozen dozen and a half a dozen dozen? 228. and one anonymous examination paper. The followl11g examples come from three famous nineteenth-century mathematicians. and the object was to visit every town once and only oncc. and sold to J. R. the famous mathematician. under which form the puzzle was known as the Icoslan Game. and yet one remained in the dish. makers of fine chess sets. It was published in London in 1859. Hamilton. They were jOl11ed by black lines along the edges.
R
w
B
. indicating the routes between them. with the names of twenty cities marked at the vertices.62
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227.

Professor de Morgan said he was x years old in the year Xl AD. Where will they meet. astrology. eight black pawns.The Puzzles
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The First Pursuit Problem The first pursuit problem appeared in a Cambridge Tripos paper of 5 January 1871. hyper-space. The figure represents a portion of a chessboard. Here is a variation: 230. and how long will it take them to catch each other? 231. One mbve consists of moving a pawn into the adjacent
a b c d e f
9
h
H G F E D C B A
•
. In subsequent editions. 232. cryptography and cyphers. When was he born?' Rouse Ball's Mathematical Recreations and Essays Rouse Ball was the original author of the famous Mathematical Recreations and Essays. and other subjects. each chasmg the next. Four dogs start from the four corners of a square field. and in the lower right corner. of side 100 yards. The first edition was published in 1892. magic squares and unicursal puzzles. The Age of Augustus de Morgan 'Writing in 1864. the recreational material became predominant. It concerned three bugs. as well as the Essays of the title on the Cambridge Mathematical Tripos. and race towards each other with constant speed of 3 yards per second. mechanical problems. and contained chapters in arithmetical and geometrical recreations. In the top left corner are eight white pawns. all the dogs starting off in a clockwise direction.

In his Recreations mathematiques Lucas discussed Sam Loyd's 'Fifteen' puzzle under the title Le Jeu du Taquin and then generalized it to consider any arrangement of squares. No diagonal or backward moves are permitted. or jumping a piece over an adjacent piece. J.64
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empty square either horizontally or vertically. Is it possible to exchange C for D and also A for B?
I
-
J
C D
-
I
A
B
J
. a rectangular circuit with two additional squares. I have a large number of stamps to the values of Sp and 17p only. into the empty square immediately beyond. Here is a simple case. again horizontally or vertically. What is the largest denomination which I cannot make up with a combination of these two different values?
The Tower of Hanoi and Other Puzzles
Edouard Lucas (1842-91) was a mathematician who studied the Fibonacci sequence and the Lucas sequence. Four pieces occupy the shaded squares as shown.
Sylvester (1814-97) sent the following puzzle to the Educational Times. a journal famous in its day for its mathematical problems and the eminent mathematicians who contributed:
233. 234. How can the black and white pawns be exchanged in the minimum number of moves?
Sylvester and the Postage Stamp Problem
J. which was named in his honour.

Train A requires to overtake train B. making use of the cul-desac line. anagramming Lucas and the Lycee SaintLouis. an exotic and faraway country that was also a French colony. Lucas's best-known invention is his 'Tower of Hanoi'. each 8 feet long. Lucas posed the first shunting problems: 235. long enough to contain only half of train B. Wishing to cross the moat. How many New York-Le Havre ships will the ship leaving Le Havre today meet during its journey to New York? 238. you pick up two planks. which was presented to the public in 1883 as the creation of Mr Claus. A garden is surrounded by a square moat of uniform width. or merely in a delayed response to the coming of the railways. Every day at noon a ship leaves Le Havre for New York and another ship leaves New York for Le Havre. which is. How would you advise the drivers?
Voie in cul-de-sac
-A B
Voie principale
236. to reach the garden. where he was then teaching.The Puzzles
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Inspired perhaps by his Jeu du Taquin. of the College of Li-Sou-Stian. Hanoi was the capital of Vietnam. The trip lasts seven days and seven nights.
. but the moat is 10 feet wide. however. Can you cross it safely?
237.

temple. the following story was published. each a cubit high and as thick as the body of a bee. but he was also a witty composer of puzzles and entertainments.66
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The year after its publication. This is the Tower of Bramah. Day and night unceasingly the priests transfer the discs from one diamond needle to another according to the fixed and immutable laws of Bramah. On one of these needles. God placed sixty-four discs of pure gold. and with a thunderclap the world will vanish.
Lewis Carroll (Reverend Charles Ludwig Dodgson) (1832-98)
Carroll is world-famous as the author of Alice in Wonderland and Alice Through the Looking Glass. When the sixty-four discs shall have been thus transferred from the needle on which at the creation God placed them to one of the other needles. The puzzle is to say how many moves are needed to transfer all sixty-four discs. as well as being. the largest disc resting on the brass plate. beneath the dome which marks the centre of the world. least importantly to
. which require that the priest on duty must not move more than one disc at a time and that he must place this disc on a needle so that there is no smaller disc below it. tower. to explain the puzzle: In the great temple at Benares. not by Lucas. says he. and Brahmins alike will crumble into dust. at the creation. rests a brass plate in which are fixed three diamond needles. and the others getting smaller and smaller up to the top one.

80 per cent an arm. Find the chance of their being the vertices of an obtuse-angled triangle. What is now the chance of drawing a white counter?' 240. in which one word is transformed into another. one letter at a time. 'I have two clocks: one doesn't go at all. a lecturer in mathematics at Christ Church. A white counter is put in. then. But we know that at London twenty-fours hours after Tuesday morning it is Wednesday morning. 75 per cent an ear. and all his writings were riddled with puns. Where. if the whole world were land we might go on tracing. never completed. 'Supposing on Tuesday. the bag shaken. word-play and logical phantasy. Tuesday morning all the way round. every 15 minutes. The second volume was called Pillow Problems.The Puzzles
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the rest of the world.' 242. 'A bag contains one counter. The Chelsea Pensioners 'If 70 per cent have lost an eye. under the general title Curiosa Mathematica. meets one coming towards him in 121 minutes: when will he be overtaken by one? 245. known to be either white or black. which proves to be white. Oxford. 85 per cent a leg: what percentage at least must have lost all four?' 244. Tuesday morning. The Two Omnibuses Omnibuses start from a certain point. in its passage round the earth. But mathematical and logical problems were only a small part of his output: he invented the word ladder. 239. and illustrated his great ability to solve problems in his head. in another hour it would be Tuesday morning at the West of England. does the day change its name?'
.' 241. travelling In both directions. as BLACK into WHITE. 'If four equilateral triangles be made the sides of a square pyramid: find the ratio which its volume has to that of a tetrahedron made of the triangles. starting on foot along with one of them. and the other loses a minute a day: which would you prefer?' 243. He planned a series of books. it is morning in London. A traveller. till in twenty-four hours we got to London again. 'Three points are taken at random on an infinite plane. and a counter drawn out.

'Two travellers spend from 3 o'clock till 9 in walking along a level road. and home again. Subtract. under the shillings the number of shillings. and under the pence the number of pounds. the daughter 1651bs. Under the pounds put the number of pence. the son 901bs. It would have been dangerous for any of them to come down if they weighed more than 15 lbs more than the contents of the lower basket. but the situation has been turned on its side. up hill 3. Is there now more brandy in the water. 'A captive Queen and her son and daughter were shut up in the top room of a very high tower. for they would do so too quick. and any number of pence under twelve. up a hill. any number of shillings under twenty. what will be the result?' 247.68
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246. This problem is described by Viscount Simon in his memoir of Lewis Carroll: 249. which exactly counterbalances a monkey which is hanging on to the other end. Outside their window was a pulley with a rope round it. and the weight 75 lbs. Suppose that the monkey begins to climb the rope. their pace on the level being 4 miles an hour.' How did they do it? The Queen weighed 195 lbs. The one basket coming down would naturally of course draw the other up. A spoonful of the brandy is transferred to the water. thus reversing the line. 'Put down any number of pounds not more than twelve. Reverse the line again.'
. One glass contains 50 spoonfuls of brandy and another glass contains 50 spoonfuls of water. Add. 'A rope is supposed to be hung over a wheel fixed to the roof of a building. A spoonful of the mixture is then transferred back to the glass of brandy. quite safely. or more water in the brandy? 250. and they also managed not to weigh less either. and down hill 6. at one end of the rope a weight is fixed. and baskets fastened at each end of the rope of equal weight. Find distance walked: also (within half an hour) time of reaching top of hill.' Query: what was Carroll's conclusion? This next puzzle is related to the river-crossing conundrum which goes back to Alcuin. 248. They managed to escape with the help of this and a weight they found in the room. and the mixture is stirred.

this dissection of a square into seven pieces from which any number of shapes can be composed. The only money he had was a half-sovereign. Could they manage it? (A half-sovereign was 10 shillings or 120 pence. a florin was 2 shillings or 24 pence. How are they composed?
. and a sixpence: so he wanted change. But a friend happened to come in. a florin.
It is no surprise that it appealed to Lewis Carroll.The Puzzles
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251. The shopman only had a crown. goes back at least as far as the middle of the eighteenth century in China. The Tangram. a fourpenny bit. 'A customer bought goods in a shop to the amount of 7s 3d. and a penny. a shilling. and a threepenny bit. a crown was 5 shillings or 60 pence. a half-crown.) 252. Here are four of the Alice figures as Tangrams. who had a double-florin.

It was developed under rather odd conditions. But by his late teens he had already produced the stunning puzzle of the riderless horses. It was an offer by Percy Williams of that amount for the best device for advertising Bergen Beach. said to me one morning. or familiar ideas in new dress. Unfortunately.' and he threw a newspaper clipping to me across the breakfast table. it came out in a bad year and did not achieve the success of some of the others. who thinks I can do anything. Not only did their lives largely overlap.70
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Loyd and Dudeney
Sam Loyd (1841-1911) and Henry Ernest Dudeney (1857-1930) will always be bracketed as the two greatest puzzle-composers of all time. 'Here's a chance. for you to earn $250. My son. It
. and a few days later I had worked out the Chinaman puzzle. Barnum bought from him and sold as 'P. I said I would take a chance at it. T. which he was about to open as a pleasure resort. Barnum's Trick Donkey'. but his parents soon moved to New York. Many years later he produced an even more amazing puzzle. co-edited by Paul Morphy. reprinted in Sam Loyd and His Chess Problems. Sam's first problem was published in the same year. Loyd was a prodigy whose chess problems alone made him famous. He was just fourteen when he started to attend a chess club with his brothers Thomas and Isaac. He considered being an engineer. 113). and by the age of sixteen he was problem editor of Chess Monthly. often shared each other's ideas (or sometimes pinched them . Sam Loyd Sam Loyd was born in Philadelphia. where he attended high school. which the circus owner and showman P. in the Strand magazine (January 1908. but they even worked together for a short period. Pop. p. of whom Isaac also became a noted problemist. Loyd had taken the old puzzle of the two dogs (problem 188) and given it a brilliant new twist. but gave up the idea when he started to make money from his puzzles.Loyd seems to have done rather more of the pinching) and competed in presenting ingenious new ideas. This is how he described the circumstances of its creation. T. the 'Get off the Earth' paradox. in an era when puzzles of all kinds were exceptionally popular and newspapers and magazines were eager to cater for their readers' enthusiasms.

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r---------------~·------------. there were thirteen Chinamen plainly pictured. fastened together so that the smaller inner one. try as you would. Scientists tried it without success. Move the inner card around a little and only twelve Chinamen remained. on my way to show the puzzle to Williams. You couldn't tell what had become of the other Chinaman. moved slightly backward and forward. He was so taken with the puzzle that
~--------------------------.---~
. producing the mystery. on a pivot. I could draw pretty well.
Cut out the three rectangles and rearrange them so that the two jockeys are riding the two horses.
consisted of two concentric pieces of cardboard. Well. As you looked at them. their artist and an old friend of mine. I stopped at the Brooklyn Eagle office to ask Anthony Fiala. but of course he knew more about it than I did. and indeed no single absolutely correct analysis was ever submitted. which was circular. to touch it up a bit for me.

The element of trickery always appealed to Loyd. and agreed to a salary of $50 a week for the puzzle column. He once edited a mechanics trade paper. then to the publisher. and before I left them they had given me an order for $250 worth of copies of the puzzle. The craze swept America. Finally they proposed that I should run a puzzle department for the Eagle. in France it was called 'Ie Jeu de Taquin' (see p. Loyd was an excellent mimic. publisher and editor. and enjoyed magic tricks and sleight of hand and telling wonderful stories with which he amused his own children . They all wanted to buy it. In Germany. which was the Rubik's Cube of the 1870s. and more.shades of Lewis Carroll. as well as writer. but I told them it was disposed of. He once gave a display of mind-reading. and produced for a number of years Sam Loyd's Puzzle Magazine. and then crossed to Europe. Deputies in the Reichstag were observed huddled over the little squares.Loyd provided the answers himself by ventriloquism. where employers posted notices forbidding employees to play with the puzzle during office hours. he produced in 1878 the '14-15' puzzle (problem 258).
. and in complete contrast. and finally to the proprietor of the paper. His son appeared to give correct answer after correct answer as Loyd held up a sequence of cards behind his back. yet his son was only miming . In between the trick donkeys and the vanishing Chinamen. He was also a self-taught wood-engraver and cartoonist. using his son as stooge.72
Penguin Book of Curious and Interesting Puzzles
he insisted on showing it first to the editor. It was said that a family servant-girl had once left their service because she heard 'voices' every day in the parlour chimney.

and also made collections of his father's work. He could see an idea from many sides at once. 'Many years ago.which also meant that Loyd was not able to patent it. in 1914. his son. Sam Loyd and his Puzzles: An Autobiographical Review. also named Sam. Fortunately. 253. risking nothing because the puzzle was impossible . He had previously compiled. that someone so creative and so brilliant at chess-problem composition should not have been a better over-the-board player or as good a mathematician as Dudeney. as well as his ability to turn his puzzles into money. the famous showman got me to prepare for him a series of prize puzzles for advertising purposes.The Puzzles
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64) and was described as a greater scourge than alcohol or tobacco. Yet his powers for rapid analysis were almost unrivalled. in his puzzles his fecund imagination. as he aptly put it. They became widely known as the Questions of the Sphinx. 'Barnum was particularly pleased with the problem of the cat and dog race. Tricks and Conundrums. The theme is reminiscent of the frog climbing out of the well. Its spread was aided by the offer of a $1000 prize. first always from the point of view of a puzzle.' It is a curiosity. "let the cat out of the bag. his ingenuity and his sense of humour were given full reign. White. This puzzle perfectly illustrates Loyd's ingenuity and humour. and a problem for psychologists to explain. on account of the large prizes offered to anyone who could master them. a very large sum in those days. exploiting the fact that they shared the same name to write as late as 1928. Loyd had first asked a New York newspaper owner to put up the prize but he refused. or. letting it be known far and wide that on a certain day of April he would give the answer and award the prizes. his friend and author of Sam Loyd and His Chess Problems. finally from the artistic aspect. for the first person to achieve a particular apparently innocuous position. for the benefit of those most concerned". when Barnum's Circus was of a truth "the greatest show on earth".
. and so Loyd offered it himself. He lacked his father's talents. then from the humorous standpoint. After his death. but possessed ample nerve. often too rapidly for him to analyse them completely. According to Alain C. The following puzzles are taken from these two books. 'Ideas came to him with great fecundity. Sam Loyd's Cyclopaedia of 5000 Puzzles. continued to produce puzzles for newspapers.

Due to weathering of the rock. one hundred feet straight-away and return. under those circumstances. engraved on a sandstone boulder.9 • •
•• 4 •
• • 4 •
• •• •
'The archaeologist is examming a completed problem in long division. The dog leaps three feet at each bound and the cat but two. and the sly reference to "letting the cat out of the bag". Can you restore the missing digits?
. and it is appropriate that this example should be a long division sum: 254. the eight legible digits provide enough information to enable you to supply the missing figures. but then she makes three leaps to his two. 'It really looks as if there should be scores of correct answers. were enough to intimate that the great showman had some funny answer up his sleeve. most of the figures are no longer legible. yet
.74
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'The wording of the puzzle was as follows: '''A trained cat and dog run a race. what are the possible outcomes of the race?" 'The fact that the answer was to be made public on the first of April.' Sam Loyd invented the cryptarithm. Now. Fortunately.

surrounded as shown by square plots of 370. as well as other would-be purchasers. so we left him arguing with katydids. How many acres are in the intenor triangular lake?
'I went to Lakewood the other day to attend an auction sale of some land. in that it gives a positive and definite answer
. The land was advertised as shown in the posters on the fence as 560 acres. This was not satisfactory to the purchasers. The three plots show the 560 acres without the lake.' 255. wished to know whether the lake area was really deducted from the land. 'The question 1 ask our puzzlists is to determine how many acres there be in that triangular lake. including a triangular lake. and shouting to the bullfrogs in the lake. only one satisfactory restoration of the problem has been suggested. 'The auctioneer guaranteed 560 acres "more or less". but since the lake was included in the sale. which in reality was a swamp. but did not make any purchases on account of a peculiar problem which developed.The Puzzles
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so far as 1 am aware. The problem is of peculiar interest to those of a mathematical turn. 116 and 74 acres. I.

' 256. England. the Governor banteringly exclaimed. ' If you know nothing about that weird relic of the early Saxons..
'Many years ago.76
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to a proposition which. Loyd. there would be a capital subject for a puzzle. Rearrange the six pieces to make the best possible picture of a horse. Curtin. when I was returning from Europe in company with Andrew G. It represents the figure of a colossal white horse. "Now. the accompanying sketch will afford an excellent idea of its appearance. engraved on the side of the mountain about a thousand feet above the level of the sea and easily seen from a distance of some fifteen miles . several hundred feet long. produces one of those ever-decreasing.. but never-ending decimal fractions."
. according to usual methods. the famous war Governor of Pennsylvania (returning from his post in Russia to seek nomination for president of the United States) we discussed the curious White Horse monument on Uffington Hill. Berkshire. After the white horse had been thoroughly discussed.

' 257. patented numerous inventions. I speedily improvised the accompanying figure of a horse. but somehow I love the old nag best as first devised. which do not require a five-dollar bill to promote and place on the market. probably not one out of a thousand. and I did modify it in the version which I afterwards published. That is all there is to it. actually mastered the puzzle. so I now present it as it actually occurred ro me. So. so it will be a capital test of the acumen of the past compared with that of the present generation ro see how many clever wits of today can solve it. but the entire world laughed for a year over the many grotesque representations of a horse that can be made with those six pieces. and puzzlers are much sharper than they used to be. In those days very few. and devoted much time and money. with my scissors and a piece of silhouette paper. 'The world has been moving rapidly during the last decade. 'It would be a simple matter to improve the parts and general form of the old horse. 'I sold over one thousand million [sic] copies of "The Pony Puzzle". How would you cut this gingerbread dog's head into two pieces of the same shape?
. then try to arrange them to make the best possible figure of a horse. 'Trace an exact copy of the figure as shown. more money is made from little things like "The Pony Puzzle". This prompts me to say that whereas I have brought out many puzzles.The Puzzles
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'Many a good puzzle idea has come from just such a tip. to my sorrow. with all its faults. Cut out the six pieces very carefully. upon the "big things".

of a distin-
. 'A prize of $1000. to bring them back to the present position in every respect except that the error in the 14 and 15 was corrected. 'People became infatuated with the puzzle and ludicrous tales are told of shopkeepers who neglected to open their stores. The fifteen blocks were arranged in the square box in regular order. has never been claimed. You see. In her anxiety to be fair and equitable in the matter. Toodles has received the present of a gingerbread dog's head and is told that she must divide it evenly with her little brother. but with the 14 and 15 reversed as shown in the above illustration. 'How many of our clever puzzlists can come to her assistance by showing how the dog's head may be divided?' 258.
'Older inhabitants of Puzzleland will remember how in the seventies I drove the entire world crazy with a little box of movable blocks which became known as the "14-15 Puzzle". Slide the numbered blocks into serial order. one at a time. although there are thousands of persons who say they performed the required feat.78
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'Here is a practical problem in simple division calculated to baffle some of our puzzlists. she wishes to discover some way to divide the cake into two pieces of equal shape and size. The puzzle consisted of moving the blocks about. offered for the first correct solution to the problem.

" said Farmer Casey. who is as old as all of us put together. and how much time they will spend over this seemingly simple problem.
1 4
5
2
3
7
11
6
10
14
8
12
9
13
15
259. Pilots are said to have wrecked their ships. An Odd Catch 'Ask your friends if they can write down five odd figures that will add up to fourteen. A famous Baltimore editor tells how he went for his noon lunch and was discovered by his frantic staff long past midnight pushing little pieces of pie around on a plate! Farmers are known to have deserted their plows. of which this is the first: 'Start again with the blocks as shown in the large illustration and move them so as to get the numbers in regular order.' 261. to say "figures" and not "numbers". Flattered by this interest shown in his family affairs. the suburban resident replied: '''My son is five times as old as my daughter." 'How old was the boy?' 260. and my wife is five times as old as the son. is celebrating her eighty-first birthday today. however. and engineers rush their trains past stations. but with the vacant square at upper-left-hand corner instead of right-hand corner. The mysterious feature of the puzzle is that none seem to be able to remember tlie sequence of moves whereby they feel sure they succeeded in solving the puzzle. It is really astonishing how engrossed most people will get. and I have taken one such instance as an illustration for the sketch. and I am twice as old as my wife.' Loyd then gives three further puzzles. Casey's Cow '''Some cows have more sense than the average man. You must be careful. '''What is the age of that boy?" asked the conductor. "MyoId brindle was standing on a bridge
.The Puzzles
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guished clergyman who stood under a street lamp all through a wintry night trying to recall the way he had performed the feat. whereas grandmother.

five feet from the middle of the bridge.80
Penguin Book of Curious and Interesting Puzzles
the other day. just twice the length of the bridge away from the nearest end of the bridge. three inches of her rear would have been caught on the bridge!" 'What is the length of the bridge and the gait of Casey's cow?' 262. Rearrange the eight pieces to form a perfect chessboard. while playing chess with the Duke of Burgundy. coming toward her at a 90-mile an hour clip. '''Without wasting a moment in idle speculation. placidly looking into the water. the cow made a dash toward the advancing train and saved herself by the narrow margin of one foot. which he wanted made into an endless piece of thirty links. and if a new endless chain could be bought for a dollar and a half. The Missing Link 'A farmer had six pieces of chain of five links each. Suddenly she spied the lightning express. how much would be saved by the cheapest method?' 263. If she had followed the human instinct of running away from the train at the same speed. If it costs eight cents to cut a link open and eighteen cents to weld it again.
'In the history of France is told an amusing story of how the Dauphin saved himself from an impending checkmate. by smashing the chessboard into eight
.

fifty miles long by fifty miles wide. given to teach a valuable rule which should be followed in the construction of puzzles of this kind. advances fifty miles at a constant rate while a courier starts at the middle of the rear and makes a complete circuit around the army and back to his starting point. I shall give our puzzlists a simple little problem suitable for summer weather. and on his return trip he will travel less than fifty miles because the rear of the army is advancing towards him. 'The smashing of the chessboard into eight pieces was the feature which always struck my youthful fancy because it might possibly contain the elements of an important problem. 'The puzzle is a simple one. he could clearly have to travel fifty miles forward and the same distance back. 'A more difficult puzzle is created by the following extension of the theme. As the army marches forward at a constant rate. Show how to put the eight pieces together to form a perfect 8 x 8 checkerboard. The restriction to eight pieces does not give scope for great difficulty or variety. It is assumed. and C can walk a mile in twenty minutes. call him A. Tandem Bicycle 'Three men wish to go forty miles on a tandem bicycle that will carry no more than two at a time while the third man is walking. a courier starts at the rear of the army.' 264. that the courier always rides at a constant speed. A square army. How far does the courier travel?' 265. It is a story often quoted by chess writers to prove that it is not always politic to play to win. B can walk a mile in fifteen minutes. He arrives back exactly at the time that the army completed an advance of fifty miles. The courier's speed is constant. By giving no two pieces the same shape. rides forward to deliver a message to the front. and has given rise to a strong line of attack in the game known as the King's gambit. and he completes his circuit just as the army completes its advance. How far altogether did the courier travel? 'If the army were stationary. other ways of doing the puzzle are prevented. he must go more than fifty miles to the front. but not feeling at liberty to depart from historical accuracy. walks at a rate of one mile in ten minutes. One man. But because the army is advancing. and the feat is much more difficult of accomplishment.The Puzzles
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pieces over the Duke's head. of course. to be found in many old puzzle books. The bicycle travels at forty miles an
. then returns to his position at the rear. 'An ancient problem. concerns an army fifty miles long.

'Someone once asked the Swiss girl how to make a Maltese cross and she replied. that they use the most efficient method of combining walking and cycling?' 266. The cutting must follow the lines ruled on the paper. assuming. of course. The white cross . the Swiss girl asks you to cut the Rag in her left hand into two pieces that will fit together to make a rectangle of five-by-six units.82
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hour regardless of which pair is riding it. A Swiss Puzzle 'This pretty Swiss miss is extremely clever at working geometrical cutting puzzles. "Pull its tai!!'"
.in the centre-of the flag is actually a hole in the paper. What is the shortest time for all three men to make the trip. She has discovered a way of cutting the piece of red wall paper in her right hand into two pieces that will fit together to form the Swiss flag she is holding in her left hand. 'For a second puzzle.

Thus. can be shown to exist between all the ancient signs and symbols. and a triangle? All five pieces must be used for each assembly. DIssecting a Cross ~It is a remarkable fact that a mysterious affinity. The Trapezoid Puzzle How can these five pieces be variously assembled to form a square. a rectangle. into four similar parts which may be regrouped to present a square. etc. 'One of these interesting transformations consists in dissecting the Greek Cross.' 268. the cross into a triangle.
. a Swastika can be changed into a square. or relationship. Work out the problem mentally before applying your scissors. a rhombus. the square into a cross.The Puzzles
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267. shown above. in that each one can be converted into another by rearrangement of its dissected parts. etc. with a central open space in the form of a small Greek Cross. a Greek Cross..

. Delicatessen Arithmetic 'Mrs Simpkins counted out the correct amount of money and said to delicatessen Louis: "Give me a pound and a half of bologna for boarders. and remarked: "It weighs toc over. since the rectangle is 5 x 13.···· ~~!11 ••••• 1I1. • . 'How much did she expend on the bologna?'
. what others had missed. A Paradoxical Puzzle This puzzle perfectly illustrates Loyd's ability to spot. the milkman.!IIIi···." said Mrs Simpkins.. 'To measure exactly two quarts of milk in each of those pails is a measuring problem pure and simple. apparently. as follows: 'Honest John. and they each demanded two quarts of milk in a hurry." 'Louis cut off a piece.: • • •
iii:~
Loyd asks for another way to reassemble the same four pieces which will lose a square." '''Then give me half of it. 270. in a familiar situation. DIE. 11iI···
•
0. .84
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269. gain an extra square.. Then along came two customers.. and the remainder of the money will buy Sc worth of pickles. This diagram shows how an 8 x 8 square can be dissected and reassembled to. Honest John 'Here is an extension of the "measuring" idea.. which you will find more elusive.' 271. but it requires considerable cleverness to achieve the desired result with the fewest number of pourings.
JiiI~I·········Ii. but minus his measuring cans. one with a five-quart pail and the other with a fourquart pail.. devoid of trick or device. says that what he doesn't know about milk is scarcely worth mentioning.. but he was nearly flabbergasted the other day when he got out on his route with his two to-galion cans full of the lacteal fluid. I first published it in t900.. In filling the orders John proved himself considerable of a puzzler.. weighed it.

" gasped the maiden. Just draw a straight line across the doughnut to show how many pieces you could produce with one straight cut. Popping the Question 'Danny went over to urge Kate to name the day. "the waiting would have been six days shorter." '''Had I received this promise yesterday. and the puzzle consists in showing how it may be cut into the least possible number of pieces to make a large eight by eight square. '''This is entirely unexpected. Whether . How do you do it?'
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272." 'Can you tell on what day of the week Danny popped the question?' 274. this particular doughnut makes an interesting puzzle.'
273. with the pattern preserved.intended or not." said Danny. "but I will marry you when the week after next is the week before last. Carving a Doughnut 'The design shows the sort of doughnut that buddies claim the Salvation Army lassies turned out "over there". Make it Square 'This design [overleaf] contains exactly sixtyfour little squares.

whereas if she had eaten twice as much breakfast food the gain would have been 11 per cent. Then she stepped on the free weighing machine and found that her weight had increased 10 per cent. What was her weight when she arrived at the food show?' A Revolutionary Rebus Loyd published hundreds of rebuses and other simple puzzles. This example is included by way of illustration. 'One of the incidents leading to the Revolutionary War. because of its ingenuity.86
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275. especially interesting to young students. is represented by that monogram. Everything Free 'A little girl visited the food show and ate seventeen different kinds of breakfast food and gathered 10 pounds of sample packages. and was described by Loyd as the best of Its kind he had ever produced and as 'the most difficult puzzle extant':
. As Loyd presented it:
276. What was the historical event?' The Longest Queen's Tour This puzzle appeared in Le Sphinx in March 1867.

There we have the idea. are dissimilar in colour. The problem is simply to 'Place the Queen on [a chessboard) and pass her over the entire sixty-four squares and back again to point of beginning in fourteen moves.The Puzzles
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277. Pieces are considered unalike if dissimilar in any respect. alike in size.' 278. It is a pretty problem and not too difficult.)
279. returning to the starting point.'
. Two pieces may be alike in form and size yet dissimilar in number or arrangement of white and black squares. These. Any move in a straight line will do. into four exactly equal parts. so that there shall be one of the Knights in each of the parts.' 280. the first step is to mark off a single white square and a single black one. Dissecting the Chessboard 'As interesting as any of the dissection puzzles based on the chessboard. is the one which asks: 'What is the maximum number of pieces into which a chessboard can be divided without any two of the pieces being exactly alike? 'Of course.' ('Straight moves' are not limited to the ordinary Queen moves in chess. The Same Again. on the lines. Knight Dissection 'Divide the chessboard [overleaf). Almost 'Pass the Queen over the centre points of all the squares in fourteen straight moves.

and I have thrown in a few similar ideas that occurred to me. bearing upon conditional positions produced from the position of the forces as arranged for actual play. From the Start. 'It will not be amiss. I give them under one heading.. but as all can be solved in less moves than intended by the authors. 'to have a little impromptu exhibition. elucidated in a sketch.88
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281. some from "Sissa". one by Max Lange. without authorship. etc. Dr Moore.'
. In the first number of the American Chess Journal Loyd introduced the series of chess puzzles based on the ordinary line-up of the pieces which has since become so famous. I find two by Breitenfeld.' he wrote.

the Queen. how can the first player mate in four moves? (b) If both parties make the same moves. for what is a puzzle but a perplexing question? And from our childhood upwards we are perpetually asking questions or trying to answer them. and for a while he collaborated with Sam Loyd.
.
Henry Dudeney
Henry Dudeney was born in the village of Mayfield in Sussex. TIt-Bits. Sam Loyd was in the same position. He enjoyed games and was a good chess player. This makes it all the more astonishing that their levels were so consistently high. Dudeney's father was also a schoolmaster. how can the first player selfmate on the eighth move? (c) Fmd how discovered checkmate can be effected in four moves. He had an uncanny knack for appealing to the public. though. like Sam Loyd. a game that might have been designed for puzzlists. (e) Find a game wherein perpetual check can be forced from the third move.ze he asserted that 'The fact is that our lives are largely spent in solving puzzles. * Dudeney was interested in the psychology of puzzles and puzzlesolving.The Puzzles
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(a) If both parties move the same moves. especially for advertising. the Weekly Dispatch and Blighty.'
* I am mdebted to Martin Gardner's Introduction to 536 Puzzles and Curious Problems for some of thiS mformatlon. and his puzzles are more mathematically sophisticated. like modern television stand-up comedy writers. Dudeney published under his own name in a variety of magazines: the Strand. Yet Dudeney was much the better mathematician. Cassell's. and entertained children with displays of magic and legerdemain. He also played croquet. but Dudeney himself did not go to college and was also a self-taught mathematician. He started composing puzzles under the pseudonym 'Sphinx'. (d) Find how a stalemate might result in ten moves. When their collaboration ended. as might be expected. he had to keep up a constant flow of ideas. without requiring any mathematics beyond the most elementary. His paternal grandfather was a self-taught mathematician and astronomer who started as a shepherd and raised himself to the position of schoolmaster in the town of Lewes. So. he was a better problemist. In the original preface to A Puzzle-l\1i. Loyd showed greater ingenuity in exploiting his puzzles.

Dudeney also supposed that puzzles had great value in training the mind. Every man to his trade. who was of the party. in Southwark [whose) host proposed that they should beguile the ride by each telling a tale to his fellow-pilgrims..' Half a century later we are more aware of the roles of insight and imagination. which are more than the exercise of logic or reason. We would put that rather differently: it is mathematics teachers today who most exploit puzzles and mathematical recreations to entice their pupils and to illuminate mathematical ideas. and said. The following puzzles are selected from The Canterbury Puzzles (1907). as shown in the illustration.90
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But he was also a man of his age. The title puzzles of The Canterbury Puzzles are a sequence of problems proposed by 'A chance-gathered company of pilgrims. and the scholar may learn from the varlet and the wise man from the fool. "Be there any among ye full wise in the true cutting of cloth? I trow not.. he was really playing off a practical joke on the company. In the same preface he remarks that 'The solving of puzzles consists merely in the employment of our reasoning faculties. The Haberdasher's Puzzle 'Many attempts were made to induce the Haberdasher. They exhibit a wealth of imagination and ingenuity. and our mental hospitals are built expressly for those unfortunate people who cannot solve puzzles. for he was quite ignorant of any answer to the puzzle that he set them. and the 'Aha!' response. As a matter of fact. then. a natural assumption in the days when many educational theorists still believed in the idea of mental training. on their way to the shrine of Saint Thomas a Becket at Canterbury. ' 282. at one of the Pilgrim's stopping-places. so Dudeney's puzzles are not limited by his own interpretations of them. in what manner this piece of cloth may be cut into four several pieces
. He produced a piece of cloth in the shape of a perfect equilateral triangle. to propound a puzzle of some kind. Fortunately. even artistry .' Elsewhere he remarks that 'The history of [mathematical puzzles) entails nothing short of the actual story of the beginnings and development of exact thinking in man. if ye can. but for a long time without success. met at the Tabard Inn. just as an artist may have a naive theory of their own art. At last. Amusements in Mathematics (1917) and Modern Puzzles (1926). he said that he would show them something that would "put their brains into a twist like unto a bellrope". later called the Talbot. Show me.

"any knave can make a riddle methinks.The Puzzles
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that may be put together to make a perfect square. as at A. The method of doing this is subtle. 1 foot from the ceiling. but it is for them that may to rede it right. that he knew no way of doing it in any number of pieces. a spider is at a point in the middle of one of the end walls. but not in four.
." saith he.' 283. "By Saint Francis. but I think the reader wlil find the problem a most interesting one. measuring 30 feet in length and 12 feet in width and height." For this he narrowly escaped a sound beating. and without turning over any piece when placing them together." 'Now some of the more learned of the company found a way of doing it in five pieces. But when they pressed the Haberdasher for the correct answer he was forced to admit. after much beating about the bush. The Spider and the Fly 'Inside a rectangular room. and a fly is on the opposite wall. 1 foot from the floor in the centre. But the curious point of the puzzle is that I have found that the feat really may be performed in so few as four pieces.

.92
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.'
. What is the shortest distance that the spider must crawl in order to reach the fly..
as shown at B.. which remains stationary? Of course the spider never drops or uses its web. but crawls fairly.
-
140
I
loc
)(1
.

and it is required to cut it into two pieces (without any waste) that will fit together and form a perfectly square flag. and so on. and. until Hendrick catches one hog and Katrun the other. so perhaps readers will be interested in the following. This would be easy enough if it were not for the four roses. in turns.' 285. as we should merely have to cut from A to B and insert the piece at the bottom of the flag. therein lies the difficulty of the puzzle. 'This you will find would be absurdly easy if the hogs moved first. Making a Flag 'A good dissection puzzle in so few as two pieces is rather a rarity. by four counters.The Puzzles
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284. and then the second player moves both pigs one square each (not diagonally).] 'The first player moves the Dutchman and his wife one square each in any direction (but not diagonally). Catching the Hogs 'In the Illustration Hendrick and Katrun are seen engaged in the exhilarating sport of attempting the capture of a couple of hogs.'
. a complete answer is afforded in the little puzzle game that I will now explain. 'Why did they fail? 'Strange as it may seem. The diagram represents a piece of bunting.' [Dudeney instructs the reader to represent the Dutchman and his wife. but this IS just what Dutch pigs will never do. on squared paper. and the two hogs. with the four roses symmetrically placed. But we are not allowed to cut through any of the roses.

each of which is no more than 9 feet long. this plan is of a ditch which is 10 feet wide. but he cannot swim. and no two pieces the same size. It was inlaid with pieces of wood. and filled with water. that is. Amongst her treasures was a casket.94
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286. The King's Jester desperately requires to cross the ditch. 287. Bridging the Ditch As Dudeney describes it. ten inches by 1I4-inch. Many young men failed. but one at length succeeded. Lady Isabel de Fitzarnulph. 'When young men sued for the hand of Lady Isabel. and the rest of the surface was exactly inlaid with pieces of wood. but the dimensions of that strip of gold. 'Sir Hugh's young kinswoman and ward. absolutely determine the size of the top of the casket. The puzzle is not an easy one. a square dissected into distinct smaller squares. was known far and wide as "Isabel the Fair". Sir Hugh promised his consent to the one who would tell him the dimensions of the top of the box from these facts alone: that there was a rectangular strip of gold. and a strip of gold ten inches long by a quarter of an inch wide. each piece being a perfect square. How can he cross the ditch to safety? Lady Isabel's Casket This puzzle is the first appearance of the idea of a 'squared square'. though Dudeney has to resort to a narrow rectangular strip to fill a portion of the surface.'
. The only equipment he can find is a heap of eight planks. the top of which was perfectly square in shape. combined with those other conditions.

and that there was nothing that 1 could use to row me across. (1) When does the fly meet B? The fly then turns towards A and continues flying backwards and forwards between A and B. it immediately turns and flies towards B. starts at noon from one end and goes throughout at 50 miles an hour. and my chance of escape seemed of a truth hopeless. The Crescent and the Cross 'When Sir Hugh's kinsman. as. which was. The Fly and the Cars A road is 300 miles long. very wide and very deep. When 1 had untied the rope and pushed off upon the water the boat lay quite still. there being no stream or current to help me. did 1 yet take the boat across the moat?' 290. Alas! I could not swim.The Puzzles
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288. came back from the Holy Land. How. But after 1 had got into it 1 did find that the oars had been taken away. then. Sir John de Collingham. A car. A. still adventuring) was now face to face with the castle moat. it would have been had 1 not espied a boat tied to a wall by a rope. Crossing the Moat 'I [the King's Jester. and at the same time another car. he brought
. When the fly meets car A. starts from the other end together with a fly travelling at 150 miles an hour. going uniformly at 100 miles an hour. doubtless. B. indeed. (2) When wiII the fly be crushed between the cars if they collide and it does not get out of the way? 289.

using every part of the crescent.111 mice. "that many years ago they were so overrun with mice that the good abbot gave orders that all the cats from the country round should be obtained to exterminate the vermin. A record was kept. as shown in the illustration. Of course we assume that all the cigars are exactly alike in every respect. yet he is able to measure any given
.. of the classical problem by Bachet concerning the weight that was broken in pieces which would then allow of any weight in pounds being determined from one pound up to a total weight of all the pieces. In the present case a man has a yard-stick from which 3 inches have been broken off. One places an ordinary cigar (flat at one end. The Cigar Puzzle 'Two men are seated at a square-topped table. 'Some of the graduation marks are also obliterated. With those restrictions you may take any dimensions you like. assuming that they each will play in the best possible manner? The size of the table top and the size of the cigar are not given. The Damaged Measure 'Here is a new puzzle that is interesting. we will say that the table must not be less than 2 feet square and the cigar not more than 4t inches long. and I shall show how the conversion from crescent to cross may be made in ten pieces. a condition being that no cigar shall touch another. [He] explained that the crescent in one banner might be cut into pieces that would exactly form the perfect cross in the other. and so on alternately. and at the end of the year it was found that every cat had killed an equal number of mice. or the second player. though it is really very different. so that it is only 33 inches in length..96
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with him a flag bearing the sign of a crescent. so that only eight of these marks are legible. and it reminds me. pointed at the other) on the table. win?' 293. and the total was exactly 1. The Riddle of St Edmondsbury '''It used to be told at St Edmondsbury. How many cats do you suppose there were?' 292. It is certainly rather curious." said Father Peter on one occasion. but in order to exclude the ridiculous answer that the table might be so diminutive as only to take one cigar. Should the first player.111. The flag was alike on both sides.' 291. Which player should succeed in placing the last cigar. then the other does the same. It was noticed thai: de Fortibus spent much time in examining this crescent and comparing it with the cross borne by the Crusaders on their own banners . so pieces may be turned over where required.

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number of inches from 1 inch up to 33 inches. Where are these marks placed?' 294. Exploring the Desert 'Nine travellers, each possessing a motorcar, meet on the eastern edge of a desert. They wish to explore the Intenor, always going due west. Each car can travel forty miles on the contents of the engine tank, which holds a gallon of petrol, and each can carry nine extra gallon tins of petrol and no more. Unopened tins can alone be transferred from car to car. What is the greatest distance to which they can enter the desert without making any depots for petrol for the return journey?' 295. A Puzzle with Pawns 'Place two pawns in the middle of the chessboard, one at Q4 and the other at KS . Now, place the remaining fourteen pawns (sixteen in all), so that no three shall be in a straight line in any possible direction.

'Note that I purposely do not say queens, because by the words "any possible direction" I go beyond attacks on diagonals. The pawns must be regarded as mere points in space - at the centres of the squares.' 296. The Game of Bandy-ball Bandy-ball, cambuc, or goff (the game so well known today by the name of golf), is of great antiquity, and was a special favourite at Solvemhall Castle. Sir Hugh de Forti-

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bus was himself a master of the game, and he once proposed this question. 'They had nine holes, 300, 250, 200, 325, 275, 350, 225, 375, and 400 yards apart. If a man could always strike the ball in a perfectly straight line and send it exactly one of two distances, so that it would either go towards the hole, pass over it, or drop into it, what would the two distances be that would carry him in the least number of strokes round the whole course? 'Two very good distances are 125 and 75, which carry you round in twenty-eight strokes, but this is not the correct answer. Can the reader get round in fewer strokes with two other distances?' 297. The Noble Demoiselle 'Seated one night in the hall of the castle, Sir Hugh desired the company to fill their cups and listen while he told the tale of his adventure as a youth in rescuing from captivity a noble demoiselle who was languishing in the dungeon of a castle belonging to his father's greatest enemy ... Sir Hugh produced a plan of the thirty-five cells in the dungeon and asked his companions to discover the particular cell that the demoiselle occupied. He said that if you started at one of the outside cells and passed through every doorway once, and once only, you were bound to end at the cell that was sought. Can you find the cell? Unless you start at the correct outside cell it is impossible to pass through all the doorways once and once only.'

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298. The Trusses of Hay 'Farmer Tomkins had five trusses of hay, which he told his man Hodge to weigh before delivering them to a customer. The stupid fellow weighed them two at a time in all possible ways, and .informed his master that the weights in pounds were 110, 112, 113, 114, 115, 116, 117, 118, 120 and 121. Now, how was Farmer Tompkins to find out from these figures how much every one of the five trusses weighed singly? The reader may at first think that he ought to be told "which pair is which pair" or something of that sort, but it IS quite unnecessary. Can you give the five correct weights?' 299. Another Joiner's Problem 'A joiner had two pieces of wood of the shapes and relative proportions shown in the diagram. He wished to cut them into as few pieces as possible so that they could be fitted together, without waste, to form a perfectly square table-top. How should he have done it? There is no necessity to give measurements, for if the smaller piece (which is half a square) be made a little too large or small, It will not affect the method of solution.'

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300. The Rook's Tour 'The puzzle is to move the single rook over the whole board, so that it shall visit every square of the board once, and only once, and end its tour on the square on which it starts. You have to do this in as few moves as possible.'

301. The Five Pennies 'Every reader knows how to pennies so that they all touch one another. Place three in a triangle, and lay the fourth penny on top in the centre. do the same with five pennies - place them so that every touch every other penny.'

place four the form of Now try to penny shall

302. De Morgan and Another 'Augustus de Morgan, the mathematician, who died in 1871, used to boast that he was x years old in the year x 2 • My living friend, jasper jenkins, wishing to improve on this, tells me that he was a 2 + b l in a4 + b 4 ; that he was 2m in the year 2m 2 ; and that he was 3n years old in the year 3n 4 • Can you give the years in which De Morgan and jenkins respectively, were born?' 303. Sum Equals Product '''This is a curious thing," a man said to me. "There are two numbers whose sum equals their product. They are 2 and 2, for if you add them or multiply them, the result is 4." Then he tripped badly, for he added, "These are, 1 find, the only two numbers that have this property." 'I asked him to write down any number, as large as he liked, and 1

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would immediately give him another that would give a like result by addition or multiplication.' What was Dudeney's method of doing this? Water, Gas and Electricity This puzzle illustrates the difficulty of deciding which puzzles were invented by Dudeney and Loyd and which were borrowed from each other, or from older sources. Loyd claimed that he invented this puzzle about 1903, Dudeney thinks otherwise. I am inclined to think that such a puzzle - of obvious practical relevance - ought to be old, but if so then it is surprising that it does not appear in the commonest Victorian puzzle books. 304. 'There are some half-dozen puzzles, as old as the hills, that are perpetually cropping up, and there is hardly a month in the year that does not bring inquiries as to their solution. Occasionally one of these, that one had thought was an extinct volcano, bursts into eruption in a surprising manner. I have received an extraordinary number of letters respecting the ancient puzzle that I have called "Water, Gas and Electricity". It is much older than electric lighting, or even gas, but the new dress brings it up to date. The puzzle is to lay on water, gas, and electricity, from W, G and E, to each of the three houses, A, Band C, without any pipe crossing another.'

305. The Six Pennies Lay six pennies on the table, and then arrange them as shown overleaf, so that a seventh would fit exactly into the central space. You are not allowed the use of a ruler or any other measuring device, just the six pennies.

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306. Placing Halfpennies 'Here is an interesting little puzzle suggested to me by Mr W. T. Whyte. Mark off on a sheet of paper a rectangular space 5 inches by 3 inches, and then find the greatest number of halfpennies that can be placed within the enclosure under the following conditions. A halfpenny is exactly an inch in diameter. Place your first halfpenny where you like, then place your second coin at exactly the distance of an inch from the first, the third an inch distance from the second, and so on. No halfpenny may touch another halfpenny or cross the boundary. Our illustration will make the matter perfectly clear. No.2 coin is an inch from No.1; No.3 an inch from No.2; No.4 an inch from No.3; but after No. 10 is placed we can go no further in this attempt. Yet several more halfpennies might have been got in. How many can the reader place?'

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307. The Bun Puzzle 'The three circles represent three buns, and it is simply required to show how these may be equally divided among four boys. The buns must be regarded as of equal thickness throughout and of equal thickness to each other. Of course, they must be cut into as few pieces as possible. To simplify it I will state the rather surprising fact that only five pieces are necessary, from which it will be seen that one boy gets his share in two pieces and the other three receive theirs in a single piece. I am aware that this statement "gives away" the puzzle, but it should not destroy its interest to those who like to discover the "reason why".' 308. The Cardboard Chain 'Can you cut this chain out of a piece of cardboard without any join whatsoever? Every link is solid, without its having been split and afterwards joined at any place. It is an interesting old puzzle that I learnt as a child, but I have no knowledge as to its inventor.'

I~l
309. The Two Horseshoes 'Why horseshoes should be c~nsidered "lucky" is one of those things which no man can understand. It is a very old superstition, and John Aubrey (1626-1700) says, "Most houses at the West End of London have a horseshoe on the threshold." In Monmouth Street there were seventeen in 1813 and seven so late as 1855. Even Lord Nelson had one nailed to the mast of the ship Victory. Today we find it more conducive to "good luck" to see that they are securely nailed on the feet of the horse we are about to drive. 'Nevertheless, so far as the horseshoe, like the Swastika and other

1 know. But the remainder of
. The Table-Top and the Stools 'I have frequently had occasion to show that the published answers to a great many of the oldest and most widely known puzzles are either quite incorrect or capable of improvement.104
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emblems that 1 have had occasion at times to deal with. each with a hand-hole in the centre. be some esoteric or lost mathematical mystery concealed in the form of a horseshoe? 1 have been looking into this matter. He instructed the carpenter to make the cuts as in the illustration and then join the eight pieces together in the manner shown. moreover. be pleased when they find the key to the mystery. prosperity. Each shoe must be cut into two pieces and all the part of the horse's hoof contained within the outline is to be used and regarded as part of the area. 'The story is told that an economical and ingenious schoolmaster once wished to convert a circular table-top. and 1 wish to draw my readers' attention to the very remarkable fact that the pair of horseshoes shown in my illustration are related in a striking and beautiful manner to the circle. that will fit together and form a perfect circle. 1 present this fact in the form of a simple problem. all different in shape. so that it may be seen how subtly this relation has been concealed for ages and ages. My readers will. So impressed was he with the ingenuity of his performance that he set the puzzle to his geometry class as a little study in dissection.' 310. has served to symbolize health. May there not. and goodwill towards men. which is the symbol of eternity. for which he had no use. we may well treat it with a certain amount of respectful interest. into seats for two oval stools. 1 propose to consider the old poser of the table-top and stools that most of my readers have probably seen in some form or another in books compiled for the recreation of childhood. 'Cut out the two horseshoes carefully round the outline and then cut them into four pieces.

'The clever youth suggested modestly to the master that the handholes were too big.The Puzzles
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the story has never been published. He therefore proposed another way of making the cuts that would get over this objection. in which some or all of the digits in a sum are deleted.a rare example of Dudeney hitting upon a popular point that Loyd missed.' Send More Money Loyd was the first inventor of the cryptarithm. zero possibly included. so it is said. This is a correct addition sum. and that a small boy might perhaps fall through them. in which each different letter stands for a different digit. because. What is the original sum?
. it was a characteristic of the principals of academies that they would never admit that they could err. and the sum has to be reconstructed. all the wood must be used. SEND MORE MONEY 311. I get my information from a descendant of the original boy who had most reason to be interested in the matter. 'Now what was the method the boy proposed? 'Can you show how the circular table-top may be cut into eight pieces that will fit together and form two oval seats for stools (each of exactly the same size and shape) and each having similar handholes of smaller dimensions than in the case shown above? Of course. For his impertinence he received such severe chastisement that he became convinced that the larger the hand-hole in the stools the more comfortable might they be. but Dudeney first replaced the mi!>sing digits with letters to make a meaningful message .he called it Verbal Arithmetic .

unknown). which sum in two directions plus.106
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The Eight Spiders Just as Dudeney and Loyd used many old puzzles. so their puzzles have been exploited by others. alas." 'At the signal given by the mother the eight spiders shot out in eight different directions at a speed of 0. Which shall we take?" came the eager query from another. Whoever reaches the goal first shall be rewarded with the largest portion of the prey. it might be expected that there would be fewer magic hexagons. more often than not. the spiders eighty inches above the centre and the fly eighty inches below. five each in each of three directions. Food being scarce. they were grumbling. were perched on the wall at one end of a rectangular room. often adding new ideas of their own. 'An honourable family of spiders. Each of you take a different path. consisting of a wise mother and eight husky youngsters. 'Suddenly one young spider shouted with glee. If Euclid could have been summoned from his grave (location. are equal? In companson to normal magic squares. along some diagonals. apart from reflections and rotations. summing in three directions. The Magic Hexagon How can the numbers 1 to 19 be placed in the cells of this hexagon so that all fifteen sums. '''You have forgotten your Euclid. without using any other means of conveyance than your God-given legs. he would have been able to show that both the hunters and the prey were in the vertical plane bisecting the two opposite walls. "Mamma! Look! There's a fly! Let's catch him and eat him!" '''There are four ways to reach the fly. The problem of the spider and the fly (problem 283 above) has been especially fruitful in variations. 312. and no solutions at all for any other size of hexagon. but found no need of attacking it since its heart had given way at the sight of enemies on all sides.65 mile per hour. owing to the Second World War.
. This one is due to Maurice Kraitchik. There are eight ways to reach the fly. Yet it is surprising that there is just one unique solution to this problem. my darling. At the end of W seconds they simultaneously converged on the fly. when an enormous fly landed unnoticed on the opposite wall. 'What are the dimensions of the room?'
•
313.

No water got into the boat while this happened. The Boat in the Bath Tommy was floating a boat in a tub of water. so that each side of the triangle sums to 20? How can the sum be made to equal
17?
. reminiscent of the 'Nuns Puzzle'. Did the level of the water rise. The boat was initially loaded with a small metal cannon. leaving the boat floating as before. fall. or stay the same. but then the cannon fell into the water and sank to the bottom.
315.The Puzzles
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314. How can the digits 1 to 9 be placed in these circles. as a result of the cannon falling overboard? Arithmorems This is the name given to a simple but elegant type of puzzle.

they met and compared the number they had counted. A Terminating Division 'In the terminating division shown below. .counting the people passing in both directions.7.
••••••• ••••••• •
Reconstruct the completed sum. it is also possible that a dot represents a digit 7. .7. . After an hour. and in the entire division the location of only seven digits 7 are known. in either direction. However.
••• ••• •• •
. . Cutting the Cube It is easy to cut a cube into twenty-seven smaller cubes by slicing it twice vertically. Counting To and Fro Mike was standing in the doorway of his house. twice horizontally and twice from back to front. however. The Axle Poser Why does the front axle of a cart usually wear out faster than the back? 317. Who had counted most? 319. counting all those he passed. . before making the next slice. the dots represent unknown digits.108
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316. .'
• • • • 7 ••
1 ••••••
7 •••••••••
1
•••
7 ••••••
•••• •••• •••• ••• •• • •
• • • • • • •
• • • • • • •
• • • • • • •
• • • • • • • • • •
• • •• •• • • • • • • • • • • • • • • ••
• •
• ••••• 7 •
· . Tom was walking up and down the road. Suppose. This makes a total of six slices. . Is it now possible to dissect the cube into twenty-seven smaller cubes in only five slices? 318.
• 7 •••••
. . that having made one or more slices you are allowed to rearrange the pieces as you choose.

a total of £29. A Leap in Age The day before yesterday I was 13 years old.The Puzzles
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320. so that each brick is immediately over the brick below. What is the volume of the sphere remaining? 324. and anyway they did not know that there had been a mistake. apologized and gave each diner £1. The Submerged Balance A lump of lead is being weighed on a balance. A. placed on the Rat bottom of a basin. When the stone and weights balance exactly. They each gave the waiter £10. and what is the date today? 323. The Cylindrical Hole A hole 6 inches long is drilled through the centre of a solid sphere. Next year I shall be old enough to get married. At this point it occurred to the waiter that £5 would not divide equally between the three. Will it stay balanced? If not. The Lost Pound Three diners on finishing their meal are presented with a bill for £30. 'This diagram [overleaf] shows a cube with a piece cut off. The diners have each now paid £9. not knowing that the waiter had rechecked the bill and found that it was only £25. which way will it tip? 326. and the waiter has £2 in his pocket. The Overhanging Bricks (2) With the same supply of bricks and a table-top. against several iron weights. The Overhanging Bricks (1) A large number of identical rectangular bricks are piled in a vertical column. Your problem is this: can you tell from this diagram if the slice ABCD could be a Rat slice? That is. Yet they originally gave the waiter £30. What is the greatest overhang that can be achieved by sliding the bricks over each other. parallel to their longest sides? 321. but the maximum projection is measured as the maximum distance from the edge of the table to the end of a brick. the balance is submerged by filling the basin with water. B. using only four bricks? The bricks may be arranged in any manner. When is my birthday. making £27 in all. which they agree to split between them. what is the greatest projection possible. keeping the other £2 for himself. Where has the missing pound gone? 325. 322. over the edge of the table. C and D lie in a plane?'
. could the points. So he returned to the table.

what happened?
. near the border. Mastering the Masters Mary was a delightful child with a precocious mterest in chess. At once. and she was ecstatic when she won one of the games.110
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327. likewise. be worth only 90 Monia cents. Who paid for the doughnut and the soft drmks? 328. leaving him with the money he started with.as delighted when her father arranged for two chess masters to visit her house. so when the government of Moria announced that in future a Monia dollar would be worth only 90 Moria cents. before you can be certain that there is a greater than 50150 chance that at least two of them have the same birthday? The surprising result was first noted by the mathematician Harold Davenport. 329. bought a 10 cent doughnut with his Monia dollar. She was even more pleased when they both agreed to playa game with her. the government of Moma retaliated by announcing that in future each Moria dollar would. so she ". Given that Mary was actually too weak to beat either of the masters in a million years. exchanged his 90 cents for a Moria dollar. He then exchanged his 90 Moria cents for a Monia dollar. The neighbouring countries of Mona and Monia were both jealous of each other's economic success. The Matching Birthdays How many people must be gathered together in the same room. crossed into Moria and bought himself a soft drink for 10 cents. a bright young spark who lived in Monia. as conclusively as she lost the other.

' I am reminded of the great mathematician G. both give a great deal of pleasure. What Colour was the Bear? A man out hunting. Taken by surprise. and the solving of them.. 'the line of approach must be as well concealed as possible. use very little else . and others. Hardy's talking in his Mathematician's Apology of 'the puzzle columns in the popular newspapers.and in my view should have reference to . What colour was the bear?
. and turned to see that the bear had not moved. by aiming due south. he took aim and shot it. He often wrote under the pseudonyms of 'Caliban' in the New Statesman. He was also editor of British Bridge World and twice captain of England at contract bridge.principles of artistry which embody an aesthetic of their own. especially inference and deduction. lies in seeking to provide this "kick". such as Dudeney or "Caliban". a humourist and the creator of the Inspector Playfair detection mysteries. and the better makers of puzzles. In his mathematical puzzles he collaborated with his friend Sydney Shovelton.' The first two problems are based on an idea which first occurs in his 'Problems for Young Mathematicians' in The Playtime Omnibus: 330. what the public wants is a little intellectual "kick". I have put a good deal of thought into the construction of problems which at first blush appear to be insoluble. spotted a bear due east.. not only the labour of working out the answer (which for many has a very slight appeal) but also the excitement of first discovering how the answer is to be arrived at? My main pleasure. Steadying himself. The invention of such exercises. as well as producing many variations in particular themes.' In Question Time he also commented on his criteria for a good puzzle: 'Does its statement involve. in constructing puzzles. and 'Dogberry' in the News Chronicle. and nothing else has quite the kick of mathematics. Nearly all their immense popularity is a tribute to the drawing power of rudimentary mathematics. H.The Puzzles
Hubert Phillips
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Hubert Phillips was a prolific composer of all kinds of puzzles who had some strikingly original ideas. In the Introduction to The Sphinx Problem Book he explained his conception of a good inferential-mathematical puzzle: in particular. since their construction can involve . he ran directly north. through inadequacy of the data.

whether his first remark is truthful or the reverse). 'Two schoolboys were playing on the toolshed roof. having realized that he was a victim of the same trick. in talking to a Muddled. who invariably lie. arriving back at her house. a White and a Muddled were sitting side by side . A Roundabout Journey Mrs Agabegyun left her house one morning.in what order is not known. Something gave way. The Three Wise Men Three Wise Men were taking a nap when a practical joker marked a cross on the forehead of each. (3) Red. some of them red and some blue. on to the Roor below. The joker then hid behind a pillar and yelled loudly. and the Muddleds. and they were precipitated. The six answers he received were: (1) Blue. 'In a certain school in Ko. the same question. Taking a card at random. Red and Blue The island of Ko is inhabited by three different races . he put to each of them in the same order. the Whites. The other's face was quite clean. At once they awoke and each started laughing at the plight of the others. he asked each of the youngsters in turn: "What colour is this?" Then. who invariably tell the truth. Finally she turned again and walked 5 miles due north.' What colour were the cards chosen by the inspector? The next puzzle has also led to many variants. taking a second card. How did he draw this conclusion?
.112
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331. (5) Blue. and walked 5 miles south. who tell the truth and lie alternately (though one cannot tell. (2) Blue. with charcoal. (6) Blue. Where does she live? Phillips also originated the 'liars and truth-tellers' theme. (4) Red. the face of one was covered with grime. which has subsequently lent itself to endless variation: 332.the Blacks. through the roof. An inspector came in carrying a number of cards. one of which follows immediately: 333. until suddenly one of them stopped laughing and felt his own forehead. •How is this to be explained?' 334. She then turned east and walked another 5 miles. Yet it was the boy with the clean face who at once went off and washed . a Black. 'When they picked themselves up.

that the bottom four rungs of the ladder were submerged. Stritebatt '''Had a good season?" I enquired of my cricketing friend. '''Fairish. The tide was rising at the rate of eighteen inches per hour . The Lodger's Bacon 'At Aspidistra House it is the landlady's custom to put the breakfast bacon on a dish before the fire. •At the end of two hours. you know." said I. I mean. Robinson and Jones. in each of my Not Out innings. 336. Mr Jones' income is £400 2s Id. The guard's namesake lives in London." '''Thirty is pretty good. And I've only been able to play on Saturdays. But we seldom had more than one in an afternoon. and the passengers with the prefix 'Mister'." said Stritebatt. And all my best scores were Not Out ones. By the way. my average would have been 35. my lowest score was 17. and the passengers included Mr Brown. Mr Robinson lived in Leeds. The Ship's Ladder 'The good ship Potiphar lay at anchor in Portsmouth Harbour. I had no two scores the same. you know. "But I was Not Out several times. the guard lived midway between Leeds and London. fireman and guard of a certain train were Brown. that each rung was two inches wide and that the rungs were eleven inches apart. and the guard's income is exactly one-third of his nearest passenger neighbour. I finished with an average of exactly 30. '''Nothmg special. apart from the blobs. Otherwise. '''Not bad." '" Any good scores?" asked I. My friend Smith worked out that if I'd scored another dozen. It originally appeared in a Civil Service examination and created so much interest that the New Statesman persuaded Phillips to set similar problems in place of their crossword and bridge columns. so the
. 'The driver. Brown beats the fireman at billiards. An interested spectator observed that a ladder was dangling from her deck. Note the class distinctions involved in addressing the workers by their surnames alone." '''For how many innings?" '''I forget. Two blobs. Stritebatt. how many rungs would be submerged?' The next problem is a slight variant only of problem 521 in Dudeney's 536 Puzzles and Curious Problems. What is the name of the engine-driver?' 337." 'What was Stritebatt's hIghest score?' 338. Mr Robinson and Mr Jones.The Puzzles
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335.

Evans likes three rashers. Grocer. from whose wife I collected most of the village gossip." (3) "Mr Baker is not the hatter. But none of them is the namesake of his own vocation. They had forgotten to tell her that. 'Three of the fanciers have birds which are darker than their owners' feathered namesakes. Each is the owner of one bird. since Mr Baker is not the baker. there would have been a whole number of rashers each." (2) "Mr Grocer is the draper. '''361. all had shared alike.114
Penguin Book of Curious and Interesting Puzzles
lodgers may help themselves as they come down in the morning. each digit in this multiplication represented a different one. Jones is very regular. 'The club has seven members. and always takes one rasher. always leaves one at least for those who follow him. 'What number does "19" represent?' 341. Brown is greedy. but Smith. And each owner is." 'Alice was disconcerted. strange to say. four statements were made to me: (1) "Mr Draper is the hatter. Falsehoods 'Messrs Draper. only
. On this particular morning." (4) "Mr Hatter is not the baker. "The answer's 519. Robinson takes his fair share of what remains when he comes down. Incidentally. 'Who is the grocer?' 340. the namesake of the bird owned by one of the others. baker and hatter. 'Alice was Disconcerted' '''What's 19 times 19?" asked the Red Queen. 'I stayed with the human namesake of Mr Crow's bird. found only half a rasher left for him. but. Moulting Feathers 'Last summer I spent a week or so in the little-known village of Moulting Feathers. 'How many rashers did Evans take?' 339.'' said the Red Queen. and takes his fair share and then half a rasher extra. Baker and Hatter are (appropriately enough) a draper. being superstitious. in Dumpshire.'' said Alice. '''Wrong. 'When 1 tried to find out who is who." But clearly there was something wrong here. 'I subsequently discovered that three of the four statements made to me are untrue. who came down last. just to make things difficult. Its social centre is the local Bird Fanciers' Club. grocer.

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two of the fanciers - Mr Dove and Mr Canary - are bachelors. 'Mr Gull's wife's sister's husband is the owner of the raven - the most popular of the seven birds. The crow, on the other hand, is much disliked; "I can't abide him," said his owner's fiancee. 'Mr Raven's bird's human namesake is the owner of the canary, while the parrot's owner's feathered namesake is owned by the human namesake of Mr Crow's bird. 'Who owns the starling?' 342. Windows 'Sir Draftover Grabbe, Chancellor of the Exchequer, conceived the unoriginal plan of imposing a tax on windows. A window having 12 square feet of glass paid a tax of £2 3s; a 24square-foot window paid £3 Is; a 48-square-foot window paid £4 17s. 'What do you suppose was the basis of the tax?' Looking-glass Zoo This puzzle was composed by the famous astrophysicist and philosopher Sir Arthur Eddington, and originally published by Hubert Phillips in Question Time: 343. 'I took some nephews and nieces to the Zoo, and we halted at a cage marked Tovus Slithius, male and female. Borogovus Mimsius, male and female. Rathus Momus, male and female. Jabberwockius Vulgaris, male and female. The eight animals were asleep in a row, and the children began to guess which was which. "That one at the end is Mr Tove." "No, no! It's Mrs Jabberwock," and so on. I suggested that they should each write down the names in order from left to right, and offered a prize to the one who got most names right. 'As the four species were easily distinguishable, no mistake would arise in pairing the animals; naturally a child who identifi"ed one animal as Mr Tove identified the other animal of the same species as Mrs Tove. 'The keeper, who consented to judge the lists, scrutinized them carefully. "Here's a queer thing. I take two of the lists, say, John's and Mary's. The animal which John supposes to be the animal which Mary supposes to be Mr Tove is the animal which Mary supposes to be the animal which John supposes to be Mrs Tove. It is just the same for every pair of lists, and for all four species.

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"'Curiouser and curiouser! Each boy supposes Mr Tove to be the animal which he supposes to be Mr Tove; but each girl supposes Mr Tove to be the animal which she supposes to be Mrs Tove. And similarly for the other animals. I mean, for instance, that the animal Mary calls Mr Tove is really Mrs Rathe, but the animal she calls Mrs Rathe is really Mrs Tove." '''It seems a little involved," I said, "but I suppose it is a remarkable coincidence. " "'Very remarkable," replied Mr Dodgson (whom I had supposed to be the keeper) "and it could not have happened if you had brought any more children." 'How many nephews and nieces were there? Was the winner a boy or a girl? And how many names did the winner get right?' 344. Wheels around Wheels How many times does a coin rotate in rolling completely about another coin, of the same size, without slipping? 345. Covering a Chessboard If two squares are removed from a chessboard, one from each end of one of the long diagonals, can the squares that remain be covered by thirty-one dominoes, each large enough to exactly cover a pair of adjacent squares? 346. Too Many Girls In a far off land where warfare had raged for many years, the number of men was too few for the number of women who wished to marry them. While nothing could be done immediately about this sorry state of affairs, the King was determined that in future there should be more boys born and fewer girls, in anticipation of the ravages of war. With this aim in mind, he decreed that every woman should cease to bear children as soon as she gave birth to her first daughter, reasoning that while there would be some families which would have only one daughter, or even one son and one daughter, there would be many with seyeral sons followed by a single daughter, producing an overall surplus of sons. Where did his ingenious scheme go wrong? 347. Forty Unfaithful Wives 'The great Sultan was very much worried about the large number of unfaithful wives among the population of his capital city. There were forty women who were openly deceiving their husbands, but, as often happens, although all these cases were a matter of common knowledge, the husbands in question were ignorant

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of their wives' behaviour. In order to punish the wretched women, the sultan issued a proclamation which permitted the husbands of unfaithful wives to kill them, provided, however, that they were quite sure of the infidelity. The proclamation did not mention either the number or the names of the wives known to be unfaithful; it merely stated that such cases were known in the city and suggested that the husbands do something about it. However, to the great surprise of the entire legislative body and the city police, no wife killings were reported on the day of the proclamation, or on the days that followed. In fact, an entire month passed without any result, and it seemed the deceived husbands just did not care to save their honour. "'0 Great Sultan," said the vizier, "shouldn't we announce the names of the forty unfaithful wives, if the husbands are too lazy to pursue the cases themselves?" '''No,'' said the Sultan. "Let us wait. My people may be lazy, but they are certainly very intelligent and wise. I am sure action will be taken very soon." 'And, indeed, on the fortieth day after the proclamation, action suddenly broke out. That single night forty women were killed, and a quick check revealed that they were the forty who were known to have been deceiving their husbands. '''I do not understand it," exclaimed the vizier. "Why did these forty wronged husbands wait such a long time to take action, and why did they all finally take it on the same day?'"

•
348. What Moves Backwards When the express from Bristol to London is thundering towards London, some parts of the train are, at one moment, moving towards Bristol. Which parts? 349. The Backwards Bicycle A bicycle is supported vertically, but is free to move forwards and backwards when the handlebars are pushed. One pedal is at its lowest point and the other is at its highest point. If a string is attached to the lower pedal, and pulled backwards, will the bicycle move forwards or backwards? 350. The Heads of Hair There are at least 50 million people living in the United Kingdom, and no human being has more than a million hairs on their head. What is the least number of inhabitants of the United Kingdom who must have, according to the information, exactly the same number of hairs on their head?

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351. Quickies (a) 'Have you ever seen anyone running along the pavement and placing their feet on the ground in this order: right foot, right foot, left foot, left foot, right, right, left, left ... ?' (b) 'By suitably placing a six-inch square over a triangle I can cover up three-quarters of the triangle. By suitably placing the triangle over the square, I can cover up to one-half of the square. What is the area of the triangle?'
to

(c) 'When is it polite

to

overtake, or pass, on the inside only?'

(d) '''Don't forget you owe me five pence!" said Fred. '''What!'' replied Tom, "Five pence isn't worth bothering about." '''All right then," said Fred, "you can give me ten pence." 'What is the logic behind Fred's reply?' 352. An Amazing Escape 'Archaeologists, more than most scientists, destroy cherished myths with every discovery they make. When they claim, however, that the Labyrinth which trapped Theseus was merely the rooms of a palace with which he was unfamiliar, they are going

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too far. They fail to appreciate that when you are being hotly pursued by a Minotaur, you must take every opportunity to turn left or right to escape the beast, however quickly you might otherwise escape in a straight line. Naturally Theseus, who entered by the south entrance to the palace, wanted to get to his beloved Ariadne, who was waiting just outside the north entrance, as quickly as possible. What was his shortest route If he was to evade the Minotaur?' 353. Speedy Gonzales The other day, travelling by London underground, I dashed on to the platform just as my train was moving out. I caught the next one, and left the exit of my destination station at exactly the same time as I would have done had I not missed the first train. Both trains travelled at the same speed, no acts of God were involved, and I didn't have to rush to make up for lost time. Explanation please?' 354. An Intimate Affair 'At an intimate little soiree given by Lady What's-her-name the other evening, each man danced with exactly three women and each woman with exactly three men. What is more, each pair of men had exactly two dancing partners in common. An admirable arrangement which pleased Lady What's-her-name no end and also gives the reader enough information to discover exactly how intimate that soiree was. How many people attended?' 355. Rice Division 'Mr and Mrs Lo Hun were poor peasant farmers, so when Mrs Lo Hun accidentally smashed the measuring bowl which she used for measuring out the rice, she was very upset. Fortunately, her husband was skilled in the traditional art of swordfighting, and she brightened considerably when he took a strong cardboard box of rectangular shape and, with the minimum necessary number of clean plane sword cuts, produced a substitute for her bowl which actually measured out one, two, three or four measures of rice, according to her choice. 'How many cuts did her husband make, and what shape was the final article?' 356. A Mon-ster Puzzle The dlustratlon overleaf shows a mon, a Japanese family crest. Janet wanted to show it in her project on Japanese history, and she was just about to cut two equal squares of gummed paper, one black and one white, into quarters and stick them down so that they overlapped in sequence, when it struck her that eight pieces might not be necessary. Indeed, they were not, and

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eventually she managed to make the mon by using fewer pieces. How many separate pieces did she use? 357. Knot these Cubes What is the shortest knot that can be tied in three dimensions using only face connected cubes? All the cubes are the same size, the knot must be continuous with no loose ends, and the cubes must be connected by complete faces. 358. These twelve matches form one square and four triangles. How can half of them be moved to form one triangle and three squares?

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359. Two Squares in One Each of these four pieces has two rightangles, so it is hardly surprising that they can be fitted together to form a square in more than one way. In how many, precisely? 360. Half-hearted Betting Major Watson, despite being short of the readies, was feeling in a good mood, so he decided to gamble with his friend Butterworth on the toss of a coin. Starting with just £1, he bet six times, each time wagering half the money he had at the time. If he won three times and lost three times, how much did he win or lose in the end? 361. Chopsticks 'A firewood merchant had a number of blocks to chop up for firewood. He chopped each block into eleven sticks. Assuming that he chopped at the average rate of forty-five strokes per minute, how many blocks would he chop up in twenty-two minutes?' 362. The Convivial Visitor '''There are only four pubs in this village," the visitor was informed, "one in each street. The village's four streets meet at the crossroads at right-angles. This street is the High Street. '''To reach the Blue Boar from the Griffin you must turn left. To reach the Dragon from the Red Lion you have to turn right." 'The visitor entered three of the pubs; he arrived at the crossroads three times during this pilgrimage, turning left the first time, going straight across the second, and turning right the third time. He spent the night at the Blue Boar. 'Which pub stands in the High Street?' 363. Siding by Siding These five locomotives have to be driven into their respective sheds, marked with their number. Moving an engine into a shed and out again, without moving any other engine in the

A Unique Number What is the unique whole number whose square and cube between them use up each of the digits 0 to 9. The Picnic Ham 'Three neighbours gave $4 each and bought a ham (without skin. would count as just one move. so for example moving A into e.
What is the area of the triangle in the centre? 365.122
Penguin Book of Curious and Interesting Puzzles
meantime. and each vertex joined to one of the points of division. as shown in the figure. out to B and into d. once each? 366. fat. The second neighbour
. Triangles within Triangles The sides of a triangle have each been divided into quarters. 364. counts as one move. One of them divided it Into three parts asserting that the weights were equal. and bones). The puzzle is to do this as efficiently as possible.

supposed to be equal. This led to a quarrel. ranging from tall and skinny. Giants and Midgets Unlike the Household Guards. corresponded to the monetary values of $3. the second one recognized only the balance of the shop. but Private Ponce claimed that he was in both the first and third platoons. There. The remainder would form the second platoon. The distances between towns are always different. and Major Mason picked out the tallest of the shortest. to short and fubsy. Major Mason's Mercenaries were a right shower. declaring that he and all the soldiers shorter than he would form the third platoon. and the smallest man in each column was ordered to step out. and properly this time. not only were there apparently no men at all in the second platoon. not forgetting the enormous Corporal Gut and the minuscule Private Git. when measured sufficientlyaccurately. From these he chose the shortest of the tall and announced that he and all those taller than him. One afternoon. respectively. $4 and $5. would form the first new platoon. Near Neighbours Trevor the travel agent has a map of Europe on which every major town is joined to the town nearest to it. What was the result? 368. Major Mason decided to divide them into three separate groups.The Puzzles
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declared that she trusted only the balance of the shop at the corner. it appeared that the parts. What is the largest number of other towns to which anyone town can be connected?
. To the Major's rage and disgust. who could parade separately without provoking the jeers of the locals. He first assembled them in a rectangular array. the middling and the small. and instructed the tallest man In each row to step out. The Major naturally flogged every man jack of them severely. the large. The men returned to their initial positions. In what way IS it possible to settle this dispute and to divide these pieces (without cutting them anew) in such a way that each woman would have to admit that she had got at least $4 worth of ham if computed according to the balance which she trusted?' 367. which gave a stilI different result. and the third only her own balance. The third partner decIded to weigh the ham on her home balance. because the first woman kept insIsting on the equality of her division. and declared that they would do it again.

Exactly nine of these statements are false. Back to the Start A billiard ball is struck without side so that it strikes all four cushions and returns to its starting position. what is the least number ·of guards that will guarantee that all the walls can be kept under observation all the time? 370. each wall being at right-angles to its adjoining walls.
374. Exactly ten of these statements are false. Will it bounce round the table for ever. Bookworm A bookworm. one in each corner. and took the trophy. Frank had indeed once again beaten Paul's average. How
Exactly one of these statements is false. Exactly six of these statements are false. A ball shoots out of one pocket at angles of 45° to the sides. The Bouncing Billiard Ball A mathematical billiard table is in the form of a rectangle with integral sides. Exactly two of these statements are false. averaging comfortably more runs per innings than Paul. to Frank's disgust. Starting from the front
. and how far does it travel? 373. and just four pockets. Almost All Lies many of them are true? 1 2 3 4 5 6 7 8 9 10 Here are ten numbered statements. Exactly seven of these statements are false. is delighted to come across the three volumes of Dr Johnson's great Dictionary of the English Language standing on a shelf. How was this possible? 371. Batty Batting Frank had an excellent first half of the season. Exactly eight of these statements are false. or end up in one of the other pockets? 372. Without knowing the precise design of the gallery. yet for the whole season. Paul was ahead. Exactly four of these statements are false. Guarding the Gallery The new art gallery has twenty walls. Lies. In what direction is it struck.124
Penguin Book of Curious and Interesting Puzzles
369. At the end of the second half of the season. Exactly three of these statements are false. feeling very hungry. Moreover he had started the second half of the season very well. and he looked forward to once again picking up the club trophy for best overall batting average. Exactly five of these statements are false.

it bores its way through to the back of the back cover of the third volume. if a move consists of sliding one coin to touch two others. Paradoxical Dice Alan. If the front and back covers of each volume are t cm thick and the pages of each volume are 7 cm thick. OH-HO How many moves are required to transform this H into the 0. without moving any of the other coins?
When you have changed the H to 0. What was surprising was that Chris's dice nevertheless consistently beat Alan's. Barry and Chris were playing at dice. and Barry's dice consistently beat Chris's.The Puzzles
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cover of the first volume. Inverted Triangle This triangle contains ten coins. how far does the bookworm bore? 375. how many moves does it take to get back from 0 to H? 376. Alan consistently beat Barry. How was this possible?
. using three fair dice which they had each marked with their own special numbers. What is the smallest number that must be moved to make the triangle point downwards?
377.

or do something else? And why?' 380. Which Contains the Beer? A grocer has six barrels of different sizes. and a second customer also bought wine. but twice as much as the first. All the windows of the car are closed tight.
O~------~--~-------------------
X
What is the distance between the original ray and the lower mirror. parallel to one of the mirrors. The Two Girlfriends John is equally devoted to his two girlfriends. Which is the beer barrel? 381. which prevents it from touching the roof of the car. The balloon is full of coal gas and is tethered by a string. Five barrels are filled with wine and only one is filled with beer. swing right. The first customer bought two barrels of wine. holding a balloon on a string. stay upright. Blackbirds 'Twice four and twenty blackbirds Were sitting in the rain. 19. How many did remain?'
382. 'The car turns left at a roundabout.20 and 31 litres. Does the balloon swing left. to which it is parallel?
379. one of whom lives uptown and the other downtown. strikes the lower mirror at right-angles at X. 18. 16. and his own
. whichever direction it is going in. and then re-emerges along its original path. Jill shot and killed a seventh part. The Obedient Ray Two mirrors are joined at a fixed angle at 0. Since all the buses run at equal intervals. He therefore decides to catch the first bus to arrive. containing 15. It bounces a number of times.126
Penguin Book of Curious and Interesting Puzzles
378. The Balloon 'Mr Tabako's little boy sits in the back seat of a closed motor-car. and a ray of light is shone into the angle between them.

On his return. Confounded Cancellation Mr Peebles believed in giving his pupils responsibility. The Cigarette Ends A tramp collecting cigarette ends from the street can make a new cigarette out of four ends. and each was simplified by Jones Minor in the same absurd . Coin Catch I have only two coins in my pocket. They add up to 15 pence and yet one of them is not a 10 pence piece. How many cigarettes can he smoke that day? 386. he found the class rolling in the a'isles with laughter.The Puzzles
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times of arrival at the road are quite random. so when he had to leave the class one day he instructed Jones Minor to go to the board and write up some simple fraction sums which the rest of the class were to do until he returned. then you can be certain that the number on the other side is even. I make the additional claim to you that if you look at the letter on one side of a card and see that it is a vowel. yet he soon finds out that he is seeing one far more often than the other. thirty-two ends.manner.
A
2
3
B
How many cards must you turn over to check whether my additional claim is correct? 385. Looking at the board he saw that the first 'sum' was written as
1~ =_
~
4
The other three sums were also fractions with numerators and denominators below 100. he looks forward to visiting each girl with equal frequency.but in this case 'correct' . Back to Back These cards are used in a simple psychological test. Every card certainly has a letter on one side and a number on the other. He collects in one morning. What were they? 384. Why? 383. What denominations are they?
.

the centre of gravity of the beer and can together will be in the centre of the can. the centre of gravity falls. after how many days did it cover exactly one half of the pond? 390. The Lily in the Pond A water lily doubles in size. Beer from a Can You are drinking beer from a can. First he put the first man in the first room and asked another man to wait there for a few minutes. and the fifth in the fourth room. As you start to drink. arrived at a hotel and asked to be put up for the night. Horseshoe Dissection How can a horseshoe be cut into separate pieces with just two cuts?
SIX
. Finally he placed the sixth man in the fifth room and went back for the seventh man. that is. but he promised to do what he could. in the area of the leaf lying on the surface of the pond. Long-playing Poser long-playing record? How many grooves are there on a standard
389. When the can is full. seven in number. by a simple example. that an irrational number raised to an irrational power need not be irrational. If it takes 30 days to cover the pond completely. A Shaking Result At a recent conference. OK? 392. He then placed the third man in the second room. but not all. and again on leaving.128
Penguin Book of Curious and Interesting Puzzles
387.' 393. Why was the number of delegates who shook hands an odd number of times necessarily even? 391. but by the time the can is empty it is back to the centre of the now-empty can. shook hands with most of the other delegates on arrival. An Irrational Number 'Show. every 24 hours. At what point did the centre of gravity reach its minimum position? 388. the fourth in the third room. who he placed in the sixth room. as near as makes no difference. The manager actually only had six rooms available. The Hotel Reception A group of travellers. most of the delegates.

' 396. Why? It may help you to know that he is extremely healthy. not smooth. I'll take one hundred and sixteen. please. walks to the bus stop and catches the bus.' Explain. however. A Price Poser 'How much will one cost?' 'Thirty pence. he gets off the bus.The Puzzles
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394. Show that after one revolution of the hoop. the point originally in contact with the girl has travelled a distance equal to the perimeter of a square circumscribing the girl's waist. 395.' 'How much will fifteen cost?' 'Sixty pence. takes the lift to the seventh floor and then walks the rest of the way. Hula Hoop 'Consider a vertical girl whose waist is circular. On the way home. walks to the tower block entrance. Around her waist rotates a hula hoop of twice its diameter.
. and takes the lift to the ground floor. and temporarily at rest. from his flat at the top of a tower block. and is not in need of exercise.' 'Thank you. Madam. Up and Down John Smith leaves home every morning.' 'That will be ninety pence.

you can certainly fry them all in 80 seconds. by doing two pieces together and then the third. How high is the top of the ladder above the floor?
. as in the figure. independently of whether the pieces consist of single pieces or of blocks of pieces already assembled. let us call the fitting together of two pieces a "move". But can you fry them more efficiently? 398. 1 metre by 1 metre. What procedures will minimize the number of moves required to solve an n-piece puzzle? What is the minimum number of moves needed?' 399. is leaning against a wall in such a way that it just touches a box. The Jigsaw Puzzle 'In assembling a jigsaw puzzle.130
Penguin Book of Curious and Interesting Puzzles
397. Three into Two You have a frying pan which will take only two slices of bread at a time. 4 metres long. and you wish to fry three slices. The Ladder and the Box A ladder. Since each slice takes 20 seconds for each side. each on both sides.

Is it possible for a polyhedron to have an odd number of vertices at which an odd number of edges meet? 402. equals the sum of the numbers of his house to the end house in the road. The Crossed Ladders Two ladders. Where does Mr Jones Live? Mr Jones has moved to a new house in a rather long street. but they have an even number of such vertices. 16 or 32. what number does Mr Jones live at? 403. The Knockout Tournament If the number of players entered for a knockout tournament is a power of 2. and has noticed that the sum of the numbers up to his own house. How wide is the passage?
401. lean across a passageway. What happens If there is a different number of entrants? In particular. then it is easy to arrange the pairings and it is obVIOUS how many matches will take place in each round. starting from 1. If the houses are numbered consecutively. for example 8. Odd Corners The regular tetrahedron and the regular dodecahedron both have vertices at which an odd number of edges meet. but excluding it. They cross at a point 8 feet above the floor. how many matches will have to he played if thirty-seven players enter a knockout?
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400. 20 and 30 feet long.

and a mixture. The Bottle and Cork A bottle and its cork cost 21 pence and the bottle costs 20 pence more than the cork. Please explain! 405. 406. please. Knotted or not Knotted? Is this piece of rope genuinely knotted. What is the cost of each? 407. Unfortunately. What is the least you need do to discover which jar is which and restore the labels to their correct jars? 408. with a slice of toast. The Breakfast Egg Mr Oval started every day with an egg. lightly boiled. yet he never bought an egg. chocolate drops. and got his one back. A Riddle Two legs sat on three legs when along came four legs and stole the one leg. Explain.132
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404. or just in a tangle? In other words. or stretch into a straight line?
. what will happen if you pull the two ends apart? Wil\. every jar has been wrongly labelled with the label that ought to have gone on one of the other jars. aniseed balls. The Mixed-up Labels You are given three boxes containing. neither borrowed nor stole his eggs and did not keep chickens. respectively. whereupon two legs picked up three legs and threw it at four legs.it tighten into a knot.

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409. which are also circular. As you can see from the figure. each ten inches in diameter? 411. Each box contains 160 cigarettes arranged In eight rows of twenty.
Your puzzle is to explain why the number of caltssons pointing in each direction. Cigarette Extras A manufacturer produces boxes of cigarettes. in the shape of two equilateral triangles edge to edge. the baggage regulations of the train company forbid any object more than 2 metres long. Balls in a Box What is the size of the smallest cubical box which will just contain four balls. 412. How does the hunter get round this rule? 413. The Long Shot A hunter travelling by train to the forest carries with him his gun. must always be equal.3 metres long. The Problem of the Calissons Calissons are a French sweet. Unfortunately.
. The Circle and Saucers 'Our table top is circular and its diameter is fifteen times the diameter of our saucers. which come packed in a hexagonal box. the calissons can be pointing in any of three directions. How many saucers can be placed on the table top so that they overlap neither each other nor the edge of the table?' 410. which is 2. In a fully packed box.

000." he declared. 415. A notebook costs more than two pencils. each choice being independent of the other. the loser pays 1 cent. the second 2 cents. High Stakes 'Mike sat down and started shuffling the cards. so that
A
8
c
The puzzle is to move coin A so that it is between coins Band C. it is nevertheless possible to get more cigarettes into the box.999 inclusive. and how many extra can be accommodated? 414. "but I'll have my revenge next week. "What stakes?" he asked. If three erasers cost more than a notebook.999. will have ten different digits? 418. let's say that two points are chosen at random on the stick. a number chosen at random between 1. '''Let's make it a gamble. and sopn. putting a few bills and some coins in the table. like this. checking his cash.01 and I'm not playing more than ten games anyway. and game followed game until at last Mike stood up.000. Pandigital Probability What IS the probability that a ten-digit number. A Moving Poser each touches the next: Place three coins in a row. and the stick is broken at those points." 'How many games had they played. "I've got only $6. without touching either B or C. "That's my last cent I've just paid you. how much does each cost?' 417. The Broken Stick (1) A stick is broken into three pieces.134
Penguin Book of Curious and Interesting Puzzles
Assuming that the cigarettes completely fill the box. How can this be done." 'So they played.999. eraser and notebook together costs $1.000 and 9. What is the probability that they will form a triangle? To make what we have in mmd a little clearer. Double up each time. "The first game.
." "'Okay." Steve replied. that is." laughed Mike. Some are Less Equal than Others 'A pencil. and which did Mike win?' 416. and three pencils cost more than four erasers.

What IS the average length of the shorter piece? 420 A Striking Clock When a grandfather clock strikes 6 o'clock. Despite this difficulty. For six consecutive birthdays the man is an integral number of times as old as his grandson. the site of the treasure can be located. Then go one third of the way to 'can't read it' and finally travel one quarter of the way to 'd1egible'. as indicated by the words in quotation marks. The Broken Stick (2) A stick IS broken into two pieces. 8 and 11 miles. How old is each at the sixth of these birthdays? 423. How many seconds elapse between the first and last strokes when it strikes midmght? 421. How? Start at town 'squiggle' and go half of the way to 'splodge'. Buried Treasure A treasure is bUrled somewhere along a straight road on which are four towns.. at random. Multiple Ages A man and his grandson have the same birthday. A map gives the following instructions for finding' the treasure. When I mentioned this interesting coinCidence to my grandfather.' What were their ages? 424. 422. the distances between them. Unfortunately. Diagonals of a Cube This figure shows two face diagonals of a cube. being 5. . What is the angle between them?
. he surprised me by saying that the same applied to him too.The Puzzles
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419. I thought that impossible . there are 15 seconds between the first and last strokes. Grandfather and Grandson 'In 1932 I was as old as the last two digits of my birth year. in sequence. the actual names of the towns have become illegible with age.

The Trisected Angle 'A confirmed angle-watcher one afternoon ruled two lines on a clock face to mark the angle then formed by the two hands.136
Penguin Book of Curious and Interesting Puzzles
How many regions of equal area can you see in
425." he replied. Find the Centre How can the centre of a circle be found. And all on credit. '''The race hasn't been run yet. No each-way nonsense for fearless Fred. 'How much did he stake on each horse?' 427. Some time later he noticed that the two hands exactly trisected the angle he had marked. you'll owe me £200. " 'And he was right. of course. 'Fred thought for a few moments and then astounded the bookie by placing a bet on each of the nine horses. accurately. '''You might as well give me my winnings now. '''That doesn't matter." said Fred. A Great Day for the Race 'Fred Bretts noticed that there were nine runners in the big race and asked his bookie what odds he was offering. Equal Areas this figure?
426. all to win. '''3-1 on Bonnie Lass." said Fred. In how short an interval could this have happened? And how soon after 3 o'clock could he have ruled his lines in order to observe the trisection in this short a time?'
. Sir. by the use of a set-square only? 428. "When it has." smiled the bookie. 4-1 on Golden Stirrup. 9--1 on Greek Hero and 39-1 the field. 7-1 on Two's a Crowd.

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429. Eight Heads and Eight Tails Lay down sixteen coins.
How many sticks are needed to make a regular hexagon? 430. can be used to make a triangle that can be picked up and handed round without falling apart.
. Popsicle Polygons The figure shows how five iced-lolly sticks.f every coin is flat on the table? 431. heads and tails alternately as shown. Two coins only. may be touched. The problem is to rearrange the coins so that those in each vertical column are alike. called popsicle sticks in the United States. Touching Three What is the smallest number of pennies that must be placed on a table for each penny to touch exactly three others.

A Simple Angle How can an angle of 30° be constructed using only an unmarked ruler? 434.
What is the only other number. with this property?
. whose opposite edges are parallel. A Moving Problem Jack and Jill are moving to a new flat and their grand piano presents a potential problem. what are the proportions of its length to its width? 436. the original number is recovered: 30
+
25 = 55
and
5SZ
=
3025. add the two parts together and square the result. Square and Add The number 3025 has the curious property that if you split it into two parts. which is shorter than your ruler? 435. is the largest possible which can be passed round the corner. it will just pass round the corridor without being tipped on its end or being disassembled. consisting of four different digits. Fortunately.138
Penguin Book of Curious and Interesting Puzzles
432. on looking down on it from above. FaIling on Edge 'How thick should a coin have to be to have a 1 in 3 chance of landing on edge?' 433.
Given that its area. Bisecting the Segment You have an unmarked ruler. How can you bisect a given line segment.

or not. Red and Black split the stakes. and if so. I borrowed some rather primitive scales from my landlady. but unfortunately one box contains duds which are all 2 gm short in weight. and whether it is too light or too heavy. 'How should I stamp my parcel?' 438. but with no use of weights. Even more unfortunately. being rather doubtful of the balance. I will give you three pounds in return. I tried weighing the parcel in the other scale-pan. show how to establish whether there is a false coin. With three weighings on an equal-arm balance. If all the other boxes contain good coins. he has forgotten which box contains the duds. 'How much do you bid. Red?' 442. which would mean ninepence. seeing a stranger. of which Jones had brought five and Smith had brought three. Watson
. they offered to share them with him. and when they felt hungry they prepared to share their sandwiches. Sharing the Sandwiches Jones and Smith were sharing a journey. Sealed Bids 'Red and Black each stakes a 5 pence piece. However. how many weighings on a weighing machine are necessary to decide which box has the duds? 439. or heavier than that. Dud Coins by the Boxful Mr Jones has plenty of coins. ten boxes of them in fact.' said Fred.' 'Done. and being uncertain whether it was under 2 pounds (or 32 ounces) in weight. weighing 40 gm each. thoroughly justifying my suspicions.' 440. Was he? 441. This gave the weight as 36 ounces. well within the sixpenny range.The Puzzles
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437. when it would go for sixpence. If the bids are equal. Striking a Balance 'Having a parcel to send by post. who turned out to be Mr Watson. 'that if you give me two pounds. the high bidder wins the stakes but pays the low bidder the amount of his low bid.' replied Jack. which it is. A Sound Bet? 'I will bet you one pound. but. eyeing their sandwiches. Now each competes for this pool by writing down a sealed bid. which looked as if one arm was longer than the other. When the bids are simultaneously revealed. The Problem of Twelve Coins 'Among twelve coins there is one at the most which has a false weight. 'The first weighing gave 28t ounces.

who was a stranger to her. wanted?' 444. except that the line isn't there. The stranger asked for Flowers. and how much did each receive? 443. A Square Chessboard 8 chessboard? How many squares are there on an 8 x
445. arranged in a row.P. 'Another complete stranger entered the bar. three full and three empty. each coin showing either head or tail.25. The barmaid asked whether he would like Flowers or I. Tricky Tumblers Here are six tumblers. Mid-point with Compass Only You are given two points which may be thought of as the ends of a line segment. Turning Tails There are eight ways to arrange three coins in a row. can you in just seven
. and Smith taking three parts. Jones taking five parts. and changing only one coin at a time. put tenpence on the counter and asked for half a pint of beer. insisting that this would not be just. Starting with three heads showing. 447. 446. after which Watson insisted on contributing £2 to the coSt of his lunch .A. But Smith objected. How can you find the mid-point of the imaginary segment using only a pair of compasses? No ruler. How did she know what the second man. Jones immediately suggested that they split the money in proportion to their contributions. Pulling a Pint ' A stranger walked into a public bar. that counts as a move. put tenpence on the counter and asked for half a pint of beer. Who was right. and no folding to get a straight line by stealth. no straight-edge are allowed. and they shared the sandwiches equally. Upon which the barmaid immediately pulled half of Flowers.140
Penguin Book of Curious and Interesting Puzzles
accepted.
UD DUU
What is the smallest number of moves needed to leave the tumblers alternately full and empty? Every time a tumbler is picked up. or 75p. or £1.

H. Across
I. This puzzle is from The Strand Problems Book by W. 1935 The first crossword appeared on 21 December 1913. though one number in the puzzle (relating to something quite different) happens to be the area in roods of the rectangular field' known as Dog's Mead. Savage. Equipped with this information and the homely items that follow. Naturally. the reader is invited to discover that jealously guarded secret. ending up with three tails uppermost? 448. Also there were 20 shillings in £1 sterling.000 crossword books had been sold. in The New York World.The Puzzles
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turns go through the entire sequence. Farmer Dunk's mother-in-law. and by the end of the first year 350.
Note: One of the 'across' numbers is the same as one of the 'downs'. Little Pigley Farm. Area of Dog'~ Mead
111
Down
square 1. T. but crosswords did not take off until 1924 when Simon and Schuster published a book of fifty puzzles. the age of Mrs Grooby. who published in The Strand Magazine. everyone jumped on to the bandwagon.
yards. and 1 rood was one quarter of an acre.
. Crossword mania erupted. who composed puzzles for John O'London's Weekly. Value 111 shillings per acre of Dog's Mead. Readers may like to know that 1 acre was 4840 square yards. and G. cross-numbers were soon to follow. This is the only case of identity. Williams.

minus 9 down. Difference in yards of length and breadth of Dog's Mead. 10. Ted. Length of tenure (in years) of Little Pigley by the Dunks. The year when Mary was born. 8. 15.
449. Perimeter in yards of Dog's Mead. The cube of Farmer's walking speed in miles per hour. 9. 4. 13.
5. See 10 down. in which no word of more than four letters will be used. the grid will obey the usual rule that the black squares do not separate any part of the puzzle completely from the remainder. x 9 down. Farmer's youngest. Number of roods in Dog's Mead x 9 down. 6. who will be twice as old as Mary next year. Fours into Nine 'This is the grid for a children's crossword. One more than sum of digits in column 2. 10. The square of Mrs Grooby's age. 7. 14. 6. Value of Dog's Mead in pounds sterling. 15 ac.
'What is the smallest number of black squares that must be filled in order to satisfy these conditions?'
. Age of Mary. 7. 10 ac. Age of Farmer Dunk's daughter. 16. Square of number of yards in breadth of Dog's Mead. Apart from this restriction. 11. 3. Age of Farmer's firstborn. Martha. Farmer Dunk's age. 12. Date (AD) when Little Pigley came into the occupation of the Dunk family. Number of minutes Farmer takes to walk 11 times round Dog's Mead.142
Penguin Book of Curious and Interesting Puzzles
2. 8.

What is the fraction? 451. What was the amount for which the cheque had been written?'
.
This represents a common fraction written as a repeating decimal. 'Agreed!' replied her mother. But can you demonstrate this by geometry? Specifically by dissecting each of these figures into the other. 4 for Starters A number which ends in the digit 4. in the shape of a triangle. Arithmetic in Pictures Anyone can see that 5' + 10' = 11' + 2'. What is the most likely final total? 453. 'Equal shares for all!' announced Lilly. A Common Libel
-
143
EVE DID
= .The Puzzles
450. How much more cake did Father have than Lilly? 454. and conversely. Mothers and Fathers First Only one large piece of cake remained..' and so saying she cut the triangular cake into five pieces. and three smaller identical pieces for the three children. all the same shape. both being equal to 125. two large and identical pieces for Father and herself. becomes 4 times larger when the 4 is removed from the end and placed at the front. until the total exceeds 12. he discovered that he had twice as much money as the cheque had been written for. After Mr Smith had spent 68 cents.TALKTALKTALKTALK . the bank clerk accidentally mistook the number of dollars for the number of cents. the tiniest. Dollars into Cents 'When Mr Smith cashed a cheque (for less than $100). 'We shall all have pieces of exactly the same shape. What is the number? 452.. Twelve Up You toss a dice with the usual numbers 1 to 6 on its faces. using of course as few pieces as possible?
455.

Red. How few strokes are required?
• •
• • • •
•
• •
• •
•
•
• •
•
458. are only a 113 marksman. without lifting your pencil off the paper? 457. and you. Sixteen Out Taking your pencil.144
Penguin Book of Curious and Interesting Puzzles
• • •
•
• •
•
•
•
456. 'Gray is a 100 per cent marksman. in a sequence of straight strokes. Recognizing the
. Each man is provided with a pistol and an unlimited supply of ammunition. the three parties have agreed to a three-way duel. without lifting your pencil from the paper and ending up at the point where you started. a firing order is to be established and followed until one survivor remains. Black and Gray. Black is successful two out of three times on the average. never having missed a buWs-eye in his shooting career. cross out all sixteen of these dots. can you cross out all nine of these dots with four straight lines. The True! 'After a mutual and irreconcilable dispute among Red. Instead of simultaneous volleys. Four through Nine Taking your pencil.

to move to the desk either directly in front or behind. honour served. and 6 a. An Objectionable Rearrangement In Class 4C there are twenty-five desks arranged in five rows of five to form a square. You can choose the three from anywhere in the line. unless his opponent arrives within the time interval and then they fight. At whom do you shoot?' 459. inverted again on the next. Arrange them in five groups of three each. each contestant comes at a random moment between 5 a.m.m. On Tuesday they had a new teacher who instructed them to each move to a new desk. they need not for example be adjacent. Matching Matches Here is a row of fifteen matches. The Hurried Duellers 'Duels in the town of Discretion are rarely fatal. What fraction of duels lead to violence?' 460.
f\f\f\f\f\f\f\
How many moves are necessary? How many moves would you need if the rules specified that four cups be inverted at each turn? 462. on the appointed day and leaves exactly five minutes later. or to the right or left of their old desk. the seconds have decIded that you will be first and Black second in the firing order. but each move must consist of inverting three at a time. with the least fuss possible. Seven Up These seven cups have to be turned the right way up. On Wednesday Peaky had returned. Oh.The Puzzles
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disparate degrees of marksmanship. and so on. and once again the teacher instructed all the students to each move to a new desk. and a cup may be inverted on one move. by repeatedly moving one match so that it jumps over three matches. That is. under the
. There. I forgot to mention that Peaky Wilson was absent on Tuesday and his desk remained unoccupied. 'Your pistol is loaded and cocked.
461.

The Professor on the Escalator 'When Professor Stamslav Slapenarski. Cutting the Cake Only Jane and her three closest friends are to cut her birthday cake. Why? 463. How many errors would you suspect remain. it turns out that only twenty errors have been spotted by both of them. What day is it?' 466.because the 'square' will then be a shallow prism. How Many Mistakes? How many mistakes are there in this sentence: 'This sentance contanes one misteak'? What is the answer to the same question for this sentence: 'Their are three misteaks in this sentence'? 468. the Polish mathematician. and that he made each trip at a constant speed. the solution is not to cut the cube into four identical square slices and use them as quarters of the square . how many pieces of cake can the four of them cut? 465. 464. If they each make one vertical cut. When their completed proofs are compared. reaching the top after taking 125 steps. The first finds thirty errors. he reached the bottom after taking fifty steps. he then ran up the same escalator. No. accused of insolence. what is the maximum number of pieces that they can cut? If not all the slices have to be vertical. which is alright because the marzipan on top makes Patrick sick anyway. and ended up in the Headmaster's study. and the second finds only twentyfour. took five steps to everyone step before). Squaring the Cube Your poser is to dissect a cube into a square using just four pieces. today will be as far from Sunday as today was from Sunday when the day before yesterday was tomorrow. A Hectic Week 'When the day after tomorrow is yesterday. one step at a time. not detected by either of them? 467.146
Penguin Book of Curious and Interesting Puzzles
same conditions as before. As an experiment. how many steps would be visible if the escalator stopped running?'
. 'Assuming that the professor went up five times as fast as he went down (that is. Two proof readers are checking two copies of the same manuscript. walked very slowly down the down-moving escalator. Peaky objected strongly.

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469. remarked that. all the same colour. by the following reasoning: Remove one horse from the set of N + 1 horses. a plan with ten streets and three kiosks. The occupants of the kiosks now wish to draw their wares from a common central depot. All Horses are the Same Colour Here is a proof that all horses are the same colour. the same colour. Where is the fallacy in this argument?
. to leave another set of N horses. while the chances of any person in the world receiving his first Nobel prize were one in several billion (the population of the world). Nobel Prizes 'On the occasion of receiving his second Nobel prize. Next.' What is the flaw in Professor Pauling's joke argument? 471. Then it follows that it is true for any set of N + 1 horses. One horse is certainly the same colour as itself. Dr LinUS Pauling. as an example. How should this be located so as to give a minimum total length for single trips to the depot from each individual kiosk? The breadths of the streets may be neglected. all N + 1 horses have the same colour. Therefore. by our assumption. replace that horse and remove a different horse. Now assume that the title statement is true of any set of N horses. Siting a Central Depot 'The street plan of a city consists only of straight streets intersecting at right-angles. the chances of receiving a second Nobel prize were one in several hundred (the total number of living people who had receIved the prize in the past) and that therefore it was less remarkable to receive one's second prize than one's first. The figure gives.1 horses in the set.'
470. the chemist. the two horses removed each have the same colour as the other N . and at an odd number of the junctions there are kiosks. to leave a set of N horses who are all. By this argument.

saw the patient and exclaimed. B had scored 14 points. Pointing to one. I can't operate!' and sent for a deputy. whatever the result of the match. There were only three sides playing and eventually they are all going to play each other once. and C had scored 9 points.' 475.148
Penguin Book of Curious and Interesting Puzzles
472. Father's Son Lord Elphick was showing his guest the family portraits. marked with dots. but that man's father is my father's son. a goal scored might make all the difference between promotion and non-promotion. A had scored 8 points. Three Teams. two inwards. 'One suggestion that has been made involves a change in the way that points are awarded. please! 473. he remarked: 'Brothers and sisters have I none. And the authorities have been thinking for a long time of methods whereby the scoring of more goals might be encouraged. The idea is that 10 points should be awarded for a win. Why do the free vertices. Father and Son Mr Smith and his son were involved in a terrible accident at the factory where they worked. 5 points for a draw. The surgeon on duty came into the operating theatre. and his son was rushed to the emergency unit of the local hospital. and prepared for immediate surgery. Explain.' Who was represented in the portniit? 474. Mr Smith was killed outright. and 1 point for each goal scored. he on a straight line?
. Four Triangles Four identical right-angled triangles have been added to a square. New Method 'What attracts people to watch football is goals being scored. 'This method was tried out on a small scale and its success can be judged from the fact that each side scored at least one goal in every match. Therefore even if you are losing 0-5 and have no hope of winning. two pointing outwards. 'That's my son. 'Find the score in each match.

and no one shook hands with the same person more than once. I asked each person.and what are its last two digits? 477. They started very slowly. How many hands did my wife shake?' 478. Three Digits What is the largest number that can be written with just three digits.The Puzzles
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476. The first race was. the one whose horse came in last being the winner. For the Love of a Good Woman Sir Pumphret and Sir Limpney both loved the Lady Isabel and resolved to have a race. The Five Couples 'My wife and I recently attended a party at which there were four other married couples. To my surprise each gave a different answer. they chose to have a loser's race. how many hands he (or she) had shaken. 'After all the handshakes were over. a farce. No one shook hands with himself (or herself) or with his (or her) spouse. Various handshakes took place. predictably. The second race was quite different. both knights racing their mounts to the finishing line. including my wife. the winner to take her hand in marriage. Knowing that Lady Isabel was opposed to all forms of competition. using no other signs or symbols at all . Why?
. wandered off the course. and never came within sight of the finishing line. then went backwards.

what is the quickest way to time the boiling of an egg for 15 minutes?' 480. How can these three hexagon stars be cut up and reassembled into one star of the same shape?
. that is perfectly self-evident. 'Yes. The problem is to dispose them so that for seven consecutive days no girl will walk with any of her school-fellows more than once.
z
When this was first shown to Professor John Edson Sweet. Professor Sweet paused for a moment and said. a famous American engineer. Y and Z. Kirkman's Schoolgirl Problem A schoolmistress is in the habit of taking her girls for a daily walk. regular dodecagon.' What was Professor Sweet's reasoning? 482. The girls are fifteen in number. Your problem is a little simpler. into pieces which assemble into one. 481.150
Penguin Book of Curious and Interesting Puzzles
479. and on each walk are arranged in five rows of three. larger. most elegantly. such that each girl might have two companions. Polygon Products The figure shows how three regular dodecagons can be dissected. It appears that XYZ is a straight line. Common Tangents Here are three circles and six common tangents which meet in pairs at three points. X. The Egg Timer 'With a 7-minute hourglass and an II-minute hourglass.

such as a 4 x 4 board.The Puzzles
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000
483. so that the tour remams unchanged when the board is given repeated quarter-turns?
. What is the smallest rectangular board on which it is possible to do a complete tour? Can you find a board on which it is possible to make a complete tour which has rotational symmetry.well. Dodecagon into Square square. either four or six squares will be omitted from your tour. the move of a knight in chess) each time. it's not that well hidden . once and only once. Because of a hidden .feature of the two shapes. on a 4 x 4 board. This
IS
a regular dodecagon and a
Your puzzle is to dissect each of them into six pieces that will reassemble to form the other. this is not as difficult as it might seem. Indeed. wherever you start. of equal area. Knight's Tour On a small board. it is not possible to start at one square and visit every square. making a knight's move (that is. 484.

and also the sum of the six vertices. pass into or through. will it be right-handed or left-handed? 487. one in each circle. everyone of the thirty-two white squares?
486. starting where he chooses. so that the sums of the numbers in every row. The Magic Hexagram Twelve circles have been placed at the vertices and intersections of this star. The Bishop's Visitation What is the smallest number of moves in which the white bishop.152
Penguin Book of Curious and Interesting Puzzles
485. that is. How can the numbers 1 to 12 be placed. are equal?
. A Handy Problem If you turn a left-handed glove inside-out. can visit.

and for purposes of the problclll wc Idc. The bookcases are positioned so that each is centred along its wall and one inch from the wall. The Two Bookcases 'A room 9 by 12 feet contains two bookcases that hold a collection of rare erotica.
c
4i FEET D
8
A
D
c
'The owner's young nephews are commg for a visit. but with Its ends reversed. and bookcase CD is 4! feet long. The bookcase~ are narrow from front to back. The bookcases are so heavy that the only way to move them IS to keep one end on the floor as a Pivot while the other end IS swung in a circular arc.The Puzzles
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488. He wishes to protect them and the books from each other by turning both bookcases around to face the wall.' Each bookcase must end up m its startmg position. What IS the 1llll111llUIll nUlllher of sWll1gs rCLJlI1red to reverse the two bookcases?'
.lltl.
9 FEET
A
IC\I
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8
W W U.c them as straight Imc segments. Bookcase A B is 8f feet long.

place the cherry into the glass. was originally designed by Sir Isaac Newton without the use of any joints or pins. J
By moving only two matches. it is claimed. 490. composed of four matches. Can you make a bridge. Put the Cherry in the Glass This diagram represents a cocktail glass. What is it?
~d
Q V W \(J C5 11 M C3
. careful study of the square will reveal the missing symbol which should go into the empty cell. over the River Cam behind Queens' College.
~. This is your chance to imitate the great man. The Mystic Square This square has occult properties. For example. using twenty-two kitchen matches? 491. The Bridge of Matches In Cambridge. is a bridge which. and a cherry. using no glue or other adhesive materials or devices.154
Penguin Book of Curious and Interesting Puzzles
489.

A
B~~--------~----------~C
. 34. 16 and 9. and you are about to raise your telescope to count the number of cannonballs along the bottom edge. 25. what is the probability that you are. Can you say how many words are in one of its books? 495. Book Words The Ruritanian National Library contains more books than any single book on its shelves contains words. How many cannonballs do the enemy possess? 493. 41. 43. to make a larger square. the lines AA' and BB' meet at X and X divides each line in the ratio 2:1. No doubt you can solve the puzzle without Hutton's instruction. Also. A' and B' are the mid-points of two sides. In particular.21. Median Mystery This is a bit of elementary geometry. and the extra one. naturally instructed his students in methods of calculating the amount of enemy ordnance. at this very moment. when you see an enemy soldier walk up with one more cannonball. 77. breathing in at least one molecule of air that was once breathed by the great mathematician Archimedes? 496. 78.The Puzzles
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492. The Master of Ordnance appears to remonstrate with him. On this occasion you observe that the enemy have assembled their cannonballs in one square pyramid. Squares into Squares? Is it possible to assemble a number of different geometrical squares. Cannonball Quiz Charles Hutton. and then to take the original square pyramid apart and build a new. using all the original balls. 494. Archimedes' Breath At a rough and ready estimate. leaving no spaces? See how close you can get in assembling squares of the following edge-lengths into a larger square: 99. he explained how to calculate the number of cannonballs in a pile from the observed number along one edge. no two of its books contain the same number of words. Professor at the Royal Military Academy and translator of Ozanam's Mathematical Recreations. triangular pyramid. 57.

Each pigeon. in the middle of a large and deep lake. The Island in the Lake The figure shows a small island.156
Penguin Book of Curious and Interesting Puzzles
A
B~~--------~----------~C
A'
Suppose that B' is moved so that it divides CA in the ratio 2: 1. A Present-able Poser Five pigeons are flying over a field in the form of an equilateral triangle. get from the shore to the island? 499. However. 2 and 2 are also two equal numbers. on which is a tree. with only a length of rope rather more than 300 yards long. who is unable to swim. as in the figure. Equal Sums It is obvious that 2 x 2 equals 2 + 2. Equal Products. Y.
•
How might a man. divide AA'? 497. On the shore is another tree. of side 100 metres. What about different numbers? Which set of different whole numbers has the same product and the same sum? 498. How will the new intersection. which is 300 yards across. as it
.

is also a lattice point? 501. They had a pde of twenty-one matches in front of them on the table and they took turns to remove up to. With four Imes only fOUL How many can he created With SIX straIght Imes?
. five points of intersection are chosen on a square grid. The Last Match More and Less were playing a simple game. Squares and Triangles How can eight matches be placed so as to form no less than two perfect squares and four tnangles? 504. but not more than. as pigeons are wont to do. How Many Triangles With three line~ only one mangle can be created. makes a small deposit. Five Points on a Lattice Five points are chosen on a square lattice. in other words.The Puzzles
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flies. Six on Five How can six matches be placed on a table so that each of the matches touches all the other five matches? 502. Less has been less successful than More. 500. Can you recommend a strategy to Less which would make him more successful than More. The loser was the person who took the last match. So far. or at least guarantee that the games were spin more or less evenly between them? 503. three matches. The figure illustrates just one possibtlity.
Why is it certain that at least one mid-point of a line joining a pair of the chosen pOints. Explain why at least one pair of these small deposits must be at most 50 metres apart.

.:sc
. each of which borders the other three.
A
~'-----::':::"". However.158
Penguin Book of Curious and Interesting Puzzles
505. Langley..
This figure shows four countries.. named after E. only three of the countries are the same shape. so that each borders the other three? Langley's Adventitious Angles This tricky problem. is famous because It is not as simple as it seems. M. Can you remedy this defect by drawing a map of four countries. A Map-colouring Probiem It is easy to demonstrate that at least four colours are needed to colour a map so that adjoining countries are differently coloured. of identical shape and size.

The Overlapping Squares Two squares are shown overleaf. If he drops off his laundry and picks up the previous week's load every Monday night. Start at any coin of your choice. 507. Choose one of the remaining heads. whose vertex angle is 20°.
508.
. one quarter of the way up the edge. Heads over Tails Layout eight pen Illes in the circle. count four as you touch four coins in succession. how many shirts must he own to keep him going?' 511. Imperfect Products 'Prove that the product of four consecutive positive integers cannot be a perfect square. and the centre of the second square lies on the right-hand edge of the first square. The Maximal Product 'What is the largest number which can be obtained as the product of positive integers which add up to 100?' 510. Repeat until all but one of the coins are tails up.' 509.Tbe Puzzles
159
506. and turn over a head. and the fourth coin must be a head until you turn it tails up. all heads up. All you have to do is to find angle BDE. Remember. DBC = 60° and ECB = 50°. A vertex of the second square lies at the centre of the first square. and turn over the fourth coin. the larger being 10 inches square and the smaller being rather smaller. to show tails up. ABC is an isosceles triangle. The Programmer's Shirts 'A neat computer programmer wears a clean shirt every day. starting with the chosen coin. count to four. you must start each time with a head.

After a few folds it is noticeably thicker. What number did I think of?
. Repeating the same action. Fold and Fold Again Taking a large rectangular piece of thin paper in your hands. each time in half. How thick is it after fifty folds? To be more precise. Delightful Discounts Buying from your favourite store you are offered a discount of 5 per cent for payment in cash. 515. Pandigital Difference I thmk of a number which contains all the digits 1 to 9. and 20 per cent because it is sale time. and find the difference between the numbers. suppose that the original sheet is one-tenth of a millimetre in thickness. Cube Formation 'What is the shortest strip of paper 1 inch wide and black on one side that can be folded to form a 1 inch cube that is black on all sides?' 513. I reverse it. so that the first digit becomes the last. The answer also contains the digits 0 to 9. 10 per cent as a long-standing customer. you fold it in half once. and then in half again. exactly once.160
Penguin Book of Curious and Interesting Puzzles
What is the area of the overlap between the squares? 512. In what order should you take these discounts in order to pay as little as possible for your purchase? 514. you fold it fifty times.

Friends and Strangers At a small dinner party. Lady Merchant enters at the gate to the left and walks along the paths to the summer house at the right-hand corner. Folding a Square from a Rectangle 'You are given a rectangle of paper . How?
519. The Square and the Triangle These five pieces can be assembled to form a square and a triangle. at least two people will have the same number of friends present .' 518. Up and Down the Garden Path Lady Merchant's garden consists of square plots of flowers surrounded by low box-hedges. What number did I think of? 517. When I square it.The Puzzles
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516. True or false? 521. the dimensions of which are unknown. there will always be either three people who are mutual friends. every evening taking. which for the purposes of this problem means a gathering of exactly six people.true or false? 520. as far as possible. Pandigital Square I think of a number which. or three guests who are mutual strangers..
. a different route.. How Many Friends? At a party (meaning any gathering of more than two people). the product contains all the 'digits from 0 to 9 exactly once each. remains the same when I turn it upside down. with paths between the plots. curiously. You are required to determine the side of a square which has an area equal to the rectangle by merely folding the paper three times.

162
Penguin Book of Curious and Interesting Puzzles
For how many successive days can she avoid repeating herself if she is always moving towards the summer house? 522. Your puzzle is to explain why every triangular number is the sum of a square number and two other triangular numbers.
o
o o
3
o o
o
o o
o
6
o
o o
o o o
10
o
o
o o
o
A further question: why is one more than every alternate triangular number also the sum of a square and two triangular numbers?
. Triangular Numbers This figure shows why the triangular numbers were given their name by the ancient Greeks.

The Puzzles
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523. just twelve little 'animals' that can be composed of six equilateral triangles fixed edge to complete edge. You will notice that two of them resemble a V and an X.
Crook
~ ~
~
Signpost
Crown
IV\/\/
Bar
Snake
~
~
Butterfly
Hexagon
00 !¥7
Yacht
. each composed of five identical squares edge to complete edge. Tick and Cross There are twelve pentominoes. coincidentally. be assembled each using nine of the pentominoes?
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524. How can an enlarged copy of each of these two figures. Animals in the Cage Just as there are twelve distinct pentominoes. three times as wide and three times as tall. which we have called a Tick and a Cross. so there are also.

000 be written as the product of two factors each of which contains no zeros at all?
. such that each piece interlocks individually with each adjacent piece? And what is the smallest polyomino tile if the condition is only that the tessellation as a whole is interlocking. like pieces of a jigsaw puzzle. even if individual pieces are not? 526. Locking Polyominoes Each of these polyominoes is composed of twenty-five small squares.000.
What is the smallest polyomino which will tile the plane.164
Penguin Book of Curious and Interesting Puzzles
Hook
Lobster
Chevron
~
&. Zero Zeros How can 1. continuing forever in every direction. and you will notice that these two pieces interlock.
Sphinx
How can these twelve animals be packed into the rhombus which is six units along each edge? 525.000. and also that they can be used to tile the complete plane.

What is the probability that both are boys? 528.well. fifty cultured pearls and two Ming jars. The Chord in a Circle If a chord is drawn at random in a circle. If she uses all the pearls. They aren't both boys. Pearls and Jars 'Mrs Tabako has fifty natural pearls. fairly easy .
L ____ _
I
. as the dotted lines show. how should she distribute them in the two jars in such a way that when Mr Tabako enters the room and picks one pearl out of either jar at random he will have the best possible chance of picking a cultured pearl?' 529.to cut this rectangle-with-a-corner-misplaced into two identical pieces.The Puzzles
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527. what is the probability that it will be longer than the side of this equilateral triangle. Reptile Repeat It is quite easy . Two Children 'I have two children. What is the probability that both children are girls?' Now suppose that I have two children of whom the elder is a boy. inscribed in the circle?
530.

' said Jane. the extra lines drawn total one half of the perimeter of the original figure in length. can be cut into four whole pieces.' and to show her disgust she tore it into two halves. each half of the original shape was a reptile. none of the pieces may be made up of smaller parts. named for obvious reasons. 531. Starting at house No. arrange the digits in ascending and descending order to form two numbers. how far will the slab move as the rollers make one complete rotation? 535.' he continued. Simple Sums Take any four-digit number. 'this shape has four sides and is made of four copies of itself. Repeat the same process with the answer. because they are not symmetrical. and it is much harder to see the answer.' Sure enough.166
Penguin Book of Curious and Interesting Puzzles
Can you. cut it into three identical pieces? Three whole pieces. Before the Invention of the Wheel A slab is being transported on three rollers. that is. Curiously. exclaiming. What was the original shape? 533. which were equally spaced. 'that's very obvious. Reproducing Reptile 'This is a reptile. 100 metres apart on a straight road. In fact. 'it can be made up from identical smaller copies of itself. all identical in shape and all the same shape as the original Sphinx.' explained Peter. How is it done?
532. The Sphinx This shape. If the circumference of each roller is 1 metre. and subtract the smaller from the larger. 'Each of these halves is a much better reptile.eventually? 534. divisible into four copies of itself. What is the result . ' 'Really. 1 he delivered
. The Most Ridiculous Route A postman with time to spare. made a point of finishing his round by walking as far as possible while visiting his last ten houses. however.

however. waiting to move forward. Conway's Solitaire Army On an infinite square grid. middle and shortest sides of the smaller are shorter than the longest. but the light in her closet has gone out. One day. however. it occurred to him that he might do 'worse' than start at No. middle and shortest sides of the larger. Bemusing Bolts Hold two identical bolts against each other. if there are stockings of seven different colours in the drawer? 539. What was it? 536. The longest. where he was always offered a cup of tea and a bun. 1 and he planned an even longer route. without overlapping its edges? 538.
Will the bolts move apart or move closer together? 537. then all the way to 9. after walking 100 x (9 + 8 + 7 + 6 + 5 + 4 + 3 + 2 + 1) = 4500 metres. Every move consists of one man jumping over an adjacent man into the empty square beyond. and so on. and rotate them around each other. 6. and ending up at No. an army of men stand behind a starting line. as in the figure. How many stockings must she take from the stocking drawer.6. ended up at No. Packing Triangles I have two triangles. respectively. vertically or diagonally. then back to 2. just as in solitaire.The Puzzles
167
its mail and then walked to to. Can I be certain that the smaller triangle can actually be placed inside the larger.)
. one larger than the other. zig-zagging up and down the road. (The jumps may be made horizontally. Choosing in the Dark Miss Golightly is getting dressed in a hurry. to ensure that she has a pair of the same colour. as if you were 'twiddling your thumbs'. which still.

pardon the pun . and you have to imagine. Your problem. of course. The pOints are all at the vertices of identical squares. shows how an army of only eight men can send one man to the third rank beyond the starting line. You are allowed.because the problem is to decide whether it is possible to draw a square on the lattice which contains exactly seventeen lattice points in its interior and no lattice points on its perimeter. as General. This is an essential point . that the points are infinitely small. of course. to choose the size of the army and to dispose your men in any manner you choose. Points in a Square This is a square lattice.168
Penguin Book of Curious and Interesting Puzzles
~
• • • • • • • •
The figure. is to decide just how far the army is able to march. which is merely illustrative. 540.
• •
•
•
•
•
• •
• • • • • • •
• • • • • • •
•
• • • • •
• •
• • • •
•
• • • •
•
• • •
•
• • •
.

by simply inserting two strokes of the pen. and walks towards it.The Puzzles
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More generally. As this curious chase continues there is no doubt that they are both circling the tree. circling also. The Squirrel and the Hunter A hunter sees a squirrel in a tree. In the days of pounds. too quickly to notice which bits go over which other bits. where N is any integer you choose? 541. are they circling each other? William James. Farthing Fiddle One of the advantages of the decimal system of coinage is that any sum of pence can instantly be converted to pounds by inserting a decimal point. the squirrel disappears round the far side of the trunk. To Knot or not to be Knotted? Glancing at this loop of string on a table.in general. but.d. there was one five-figure quantity of farthings which could be converted into £. the famous psychologist. As he does so. What is your pragmatic response? 542. you idly ask yourself whether it is likely to be knotted? What is the answer?
. is there always a square which contains exactly N lattice points in its interior. and a farthing is a quarter of one penny. 543.s. What was it? You may recall that £1 = 20 shillings = 240 pence. posed this problem in his book Pragmatism. and as the hunter circles the tree the squirrel keeps out of sight on the other side. However. shillings and pence this was not possible .

is picked up each day at the train station at exactly 5 o'clock. whereupon he went back downstream to retrieve his paddle. Here is a puzzle and the solution as published in a book which shall be nameless. The chauffeur drove him the rest of the way home. which as you know is a large building in the form of a regular pentagon.' How much did each turkey cost? 545. Since It is equally likely that the spy will be at either spot. The Lost Paddle 'A man went upstream from his dock In a motorboat. how long will it take to pass completely through the tunnel? 547. An Express Problem An express train takes 3 seconds to enter a tunnel which is 1 km long. The Chauffeur Problem 'Mr Smith.170
Penguin Book of Curious and Interesting Puzzles
544.9-". If it is travelling at 120 km an hour. the probability is one-half. and caught up to it directly opposite his dock. Spot the boob! A spy is watching the Pentagon. Spot the Blunder Puzzle-solvers must be wide awake to solutions which seem to be solid as a rock but actually contain large holes. getting him there 20 minutes earlier than usual. Again he met the chauffeur and rode the rest of the way with him.
. the other will see three. How much ahead of usual were they this time?' 546. and again began walking home. What is the chance that he or she can see three sides of it? This is the offered solution: Imagine another spy at an equal distance away.30 train. The first and last digits were illegible. Mr Smith arrived unexpectedly on the 4. As he passed under a bridge one mile from the dock his emergency paddle fell overboard. Eventually he met the chauffeur driving to the station to get him. If one spy can see two Sides. 'On another day. from a distance with powerful binoculars. exactly opposite to the first spy. what was the rate of the current of the river?' 548. Cooked Turkey 'An old invoice showed that seventy-two turkeys had been purchased for "-67. a commuter. One day he arrived unannounced on the 4 o'clock train and began to walk home. a loss which he did not discover until 10 minutes later. If he travelled at constant water speed and lost no appreciable time turning round.

30 and the shorter one at 6.The Puzzles
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549. How long was each candle originally?' 550. it proves to be impossible to bring them into line with all the A's the same way up. With these conditions. The longer one burned out at 10.o line.OO. and the arrangement finally achieved has the plan view shown in the second figure. one of which was one inch longer than the other. The Burning Candles 'On Christmas Eve two candles.30. and the shorter one at to.30 they were both the same length.
A A A A
'The boxes are to be brought in. How long did it take for them to pass each other?
. The Heavy Boxes 'Five equal cubical boxes. and a train of the same length in the other direction takes 4 seconds.00. stand together as in the first figure. The longer one was lighted at 4. At 8. each with an A on its top side. but they are so heavy that they can be moved only by tipping them over about an edge. were lighted. Passing Trains A man standing on a platform notes that a train going in one direction takes 3 seconds to pass him. Which box was orIginally in the middle?'
A
A
A
A
A
551.

If this puzzle consisted of statements 1 to 5 only. one of whom always lies and one of whom is impeccably honest. 6. 4. the prisoner has been given a last chance. Short-list '1. 3. Which statements are true?'
. The Prisoner's Dilemma Because he is deemed to be a foolish man who has allowed himself to be led into crime by his companions. There are more true statements than false. 5. to be put to one of the warders. 2. then the answer to the following question would still be the same. one of which leads to freedom and the other to a long sentence. How can he discover which is the door to freedom? 554. There are no two consecutive true statements. He is shown two doors in the courtyard. The number of the second true statement added to the number of the first false statement gives the number of a statement which is true. but he does not know which is which. Each is guarded by a warder. The number of the first true statement here added to the number of the second false statement gives the number of a statement which is true. He is allowed one question. Triangles in a Triangle this figure?
553. There are at most three false statements.172
Penguin Book of Curious and Interesting Puzzles
How many triangles can be counted in
552.

the corners of rectangle R are such a set. What are the other two sets?'
.and seven paper disks have been tossed on a table. Each corner of a rectangle and each spot where edges intersect makes a point. Concyclic Points 'Five paper rectangles . How can the volume of this common solid be calculated without the use of any calculus? 556.The Puzzles
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555. The problem is to find three sets of four concyclic points: four points that can be shown to lie on a circle. has four identical curved surfaces. They lie as shown in the figure. For example. because the corners of any rectangle lie in a circle.one with a corner torn off . The Crossed Cylinders Two identical cylinders are placed so that their axes cross at right-angles and their common volume.

hot cross buns. in order to reach this position with White to play?
.
What is the smallest number of moves that could have been played. How many children were there if they were all treated alike and if there was only one way in which to purchase the buns?' 558. hot cross buns. the position after Black has just played. yet it is. hot cross buns. Hot Cross Buns
Hot cross buns. I had as many daughters as I have sons So I gave them seven pennies To buy their hot cross buns. three a penny. two a penny. White to Play This looks like the position after White has made a rather unusual first move. One a penny.174
Penguin Book of Curious and Interesting Puzzles
'The hot cross bun man cried:
557. in fact. If your daughters don't like them Give them to your sons! Two a penny.

They have got into the habit lately of going around in threes." 'Find the tribes to which A. but I have an excellent clock. in such a way that when the opposite edges are joined. but lies. Then John says. it is always the most important part of life on the island to discover to which tribe people belong. 11." B: "A is a Pukka. 'As the reader can imagine. the Pukkas. C is Silent 'On the Island of Imperfection there are three tribes.The Puzzles
175
559. John then asks. 17. and again he lies. to have the same sister? 562. who never tell the truth. which I occasIOnally forget to wind. The Same Sister Is it possible for two men who are completely unrelated to each other. John asks. and went back and set my clock. A cylinder can be 'squared' with the usc of only ten square pieces. passed the evening in listening to a radio concert programme. Tom's House 'John is trying to find out where Tom lives. "Is the number a multiple of 4?" Again Tom answers. who make statements which are alternately true and false. Band C belong. but the other two spoke as follows: A: "C is a Pukka. The Unwound Clock 'I have no watch. "Is it greater than 50?" and Tom answers. one from each tribe. 15. who always tell the truth. 'C did not make my self-appointed job as a detective any easier by being silent. On a recent visit I was doing some work on this with three inhabitants whom I shall call A. How could I do this without knowing beforehand the length of the trip?' 560. Once when this happened I went to the house of a friend. Finally John asks "Is the first digit 3?" After Tom has replied (truthfully or not we do not know!) John tells him the number. 25.' 563. the Wotta-Woppas. He is wrong! What was the number of Tom's house?' 561. "Is it a perfect square?" Tom answers and this time he tells the truth. it forms a 'squared' cylinder?
. 13. and the Shilli-Shallas. and I am glad to say that these three were no exception. and all he knows is that it is in a street where the houses are numbered from 8 to 100 (inclusive). How can squares of edges 30. 27. Band C. 8. or false and true. 3 and 2 be fitted together to fill the space between two parallel lines.

"
'Which of the three .A. II. Babs is not the murderess. We ended up having equal amounts of money.
Which one of the three women was the murderess?' 566. Speaking of Bets 1. glancing at the menu. B won from C as much as B then had left. each letter represents a different digit:
ABCDE x F
GGGGGG Which of the ten digits does G represent? 567.is the speaker?' 565. 'Ben owns a used car lot. 'Each of the following three statements was made by one of the three women: 1.176
Penguin Book of Curious and Interesting P.
564. Only the innocent woman told the truth. 2. The Wheels of Commerce '''How's the motor business?" asked Bob. His cars are good. I began with 50 cents. Next. named Anna.
I." '''That's dandy!" commented Bob.
First. Anna is not the accomplice. Finally. B or C . 3. C won from A as much as C then had left. The innocent woman made at least one of these statements. 3. The Murderess 'Three women. the second was an accomplice in the murder." he replied. "I was talking to Stan Logan
. were questioned about the murder of Dana. and the third was innocent of any involvement in the murder. "but sales have picked up again. 5. Other dealers come and go. III. Six Gs In the multiplication problem below.uzzles
' "The three of us made some bets. A won from B as much as A had originally. Babs and Cora. his guarantee means just what it says. "Not too bright around Christmas. One of the three women committed the murder. 4. but Ben keeps right on selling. 2. his prices are right. Each statement refers to a woman other than the speaker. Cora is not the innocent one.

The Puzzles
177
down on Wardie and Myrtle yesterday." 'Ben smiled. multiplied by the difference between the second and third weeks. "The difference between the numbers we sold in the first and second weeks. which it was reasonable to suppose had been randomly distributed in the material when it was laid down on the bed of some ancient lake. they would have been forced together in that vertical direction. had a bright idea." 'The shapely waitress leaned over his friend just then to take their order. and she no longer knew in what orientation they had been found." '''What's that in actual numbers?" asked Bob. "A lot of them are having a tough time. as the sediment was vertically compressed into rock. We've done well so far this month ." He thought a moment." he said. and Bob rather lost interest in car sales. And here's something to amuse yourself with. and examination of their present distribution would show what that direction had been! He explained his idea enthusiastically. But how many cars would you say Ben sold in the third week?' 568. but Professor Mesozoic thought for a moment. Professor Mesozoic. "but we sold fifty-six cars the first three weeks. and then rejected his idea. She had lost her notes on the samples of sedimentary rock that she had collected. Why?
. comes to the same as the number we sold the first week. '''I'm not sure about the last few days. Then her assistant. who's a great one for facts. had a problem. He's hardly sold a car this year. Within the rock were numerous tiny specks of a mineral. "but maybe I'm lucky.each week more sales than the previous week. Slatebed." replied Ben. the famous geologist. Subsequently.

The difference between each successive share is 37. We lack such
. [Peet. either a or b is even.55 or 37t\. 69]
10. Each term is then multiplied by Ii to change the 60 into 100. 1945. In this case the scribe guessed 16 immediately. The Babylonians could now find suitable values of p and q by referring to the standard reciprocal tables which they used for multiplication anyway. 46] 14. 3. without any further calculation. [Peet. p. which has the property that the first two terms sum to one-seventh of the last three. Then there are integers p and q such that a = 2pq. [Eves. Next the scribe calculates that 1 + 2/3 + 112 + 117 is 97/42 so that the multiplier required is 2/97. 1976. 11 + lOB + 20 + 291 + 38i = 100. 1923.24. The scribe first artificially constructs the series 1 + 6t + 12 + 17t + 23 = 60. The third leg is length 0. such that a' + b' = c'. 78] 11. and then adjusted by multiplying by a suitable factor. It is 1156 + 11679 + 11776.0. tOOO = 10' + 30'.30 . 53] 13. 1923. 161] 12. 46] 15.182
Penguin Book of Curious and Interesting Puzzles
be roughly correct was chosen. p. The evidence that the Babylonians used this formula is simple: the values of p and q which fit the numbers on Plimpton 322 are all socalled 'regular' numbers whose factors are powers of only 2. 1976.6 = 0. The ratio cia = t(plq + qlp).30 and one leg 0. The distance is the third side of a right-angled triangle with hypotenuse 0. suppose that a is even. b = p' . p. p. 1972. [Gillings. which the scribe could read off from the earlier table in the Rhind papyrus of fractions 2/n.18. p. which falls short of 37 by 2/42. Given that a. 62 + 82 = 100.q' and c = p2 + qZ. [Neugebauer and Sachs. [Eves. p. and got the wrong but close answer 36 + 2/3 + 114 + 1128. band c are integers. and 5 or products of such powers. This problem was also solved by false position.

1965] 17.13x
= 28938G
x. let G = 4657. Then the distance SPT equals the distance S'PT and the latter will be a mimmum when S'PT is a straight line.893.514 dappled bulls 7.060 yellow bulls 4.820 5.358.640. Dover.206. Heron used exactly the same argument by reflection to conclude that when light is reflected. p.149.
18.llt
= 9801G
42t . 4657 must divide G. we obtain as the smallest solution: 10. the angles of incidence and reflection are equal.213
[Archimedes' cattle problem. Suppose that Mary alms for point P on the river bank. as taken from T. n.482 white bulls black bulls 7.439. y. This solution follows Dome.515. The Works of Archimedes with the Method of Archimedes. Since it divides none of the coefficients on the right. As
.366. This is one of the earliest solutions to an extremal problem. L.246 3. Heath. z and t and
These equations are solved for the four unknowns we obtain: 4657 x
=
n06360G
4657y 4657t
=
=
4893246G 5439213G
4657z = 3515820G
m which the number 4657 is prime.387 white cows black cows dappled cows yellow cows 7.184
Penguin Book of Curious and Interesting Puzzles 30z . Taking the simplest case. Reflect Mary's original position in the line of the river bank.d. It follows that P IS the point such that SP and PT make the same angle with the line of the river. 319.360 4..

1980.' [Sandford.y are the sides of two such rectangles. if u. 505]
. Band C do the whole work in 10. 15 and 30 days. approximately. and my left eye fills 1112 in 6 hours. or 47 minutes and 13 seconds.' 19. Then. respectively.The Solutions
185
s·
one Greek commentator remarked on Heron's solution. 'for Nature does nothing in vain nor labours in vain. 21. p.v and x. 216] 20. and my foot 1116. where the Greeks might have taken it to be 12]. [Thomas. p. Suppose that the given ratio is n rather than 3. So the jar will be filled once in 6 x 48/61 hours. Thus all four fill the jar 1 + 118 + 1112 + 1116 = Iii times in 6 hours. 1930. A. the equations can be written:
u
+
v = n(x
+
y)
xy
=
nuv
and Heron's solution runs parallel to the general solution:
x=2n 3 -1
u = n(4n 3
-
2)
y = 2n 3 v=n
which leads Heron to his solution: the rectangles are 53 x 54 and 318 x 3. expressing a view which has haunted and inspired scientists ever since. 'My right eye fills 118 jar in 6 hours [taking a day to be 24 hours.

[The quotation from Xylander is from Ore. This is his solution. and that second = X + 6. There is no very simple solution to this problem. 'Therefore x = 20 and the numbers are 80. [Thomas. for example.'
28. so. It was left to later Indian mathematicians to find more general solutions. and the area is 210. 'Therefore product of the first and second = Xl + 6x. 'But first + third = square. 195) 26. we obtain x = 1 and the numbers are 1. 8 and 10 is another solution. 9. which illustrates very well his methods. 12. 6. 1948. and therefore the third 2x + 1. which tend to simplify the problem and produce one or a few solutions. 64 satisfy the condition. 25. sum of first and second Xl. 'By the two remaining conditions. the second.4andl1.9. 1980.21 and 29. 6x + 1 = square = 121. 20. Diophantos's answer is 1. and 9 the third number. 320.1)'. 4 and 5 is the simplest solution.7 and 9. 'Therefore the first = 4x. and the second = Xl . 'Therefore we have to find two squares differing by 48. p. The first number is 98.
Xl
'Take a square and subtract part of it for the third number. say. that is. let sum of second and third be (x . 507-8) 23. 'The squares 16. Equating these squares to the respective expressions. pp. let first = x. lOx + 54 and lOx + 6 are both squares. 94. 7. The sides are 20.4x. 41. this is easy and can be done in an infinite number of ways. but it is the ratios of the numbers which are important. This is Diophantos's solution: 'Let the sum of all three be Xl + 2x + 1. let + 6x + 9 be one of the sums. 24.7.'
.186
Penguin Book of Curious and Interesting Puzzles
22. 27.

z.m)l. Thus we have to divide 721 into two parts such that t of one part plus 1of the other = 1H.60 < 8x. 48-9] 30. and therefore m is less than 21. and therefore m is not less than 19.
Therefore the number of five-drachma measures = 79/12. (1) Since x' > 5x + 60. Let the first part be 5z.60) and <t(x1 . (2) x' < 8x + 60 or Xl = 8x + some number less than 60. whence x is not less than 11. We cannot have a real solution of this unless x > 1(x1 . or second part = 92 . Hence we put m = 20. therefore Xl . so that Xl .60 = (x . Now (from above) x = (m' + 60)/2m. Xl = 5x + a number greater than 60. Therefore 11 < x < 12. and z = 79/12. therefore 22m < m 1 + 60 < 24m. and reflect the shape in the shore-line. This is a circle.60).8z. 1968.The Solutions
187
29. the total price.20)'.60 was the price paid. Therefore i (second part) = It! . Therefore 5x < Xl . so that x = lIt. Imagine that it takes the form in the figure. for double the length of string. has to be divided into two parts such that t of one + 1of the other = x. x' = 1321.8z = 721.' [Midonick.60 = 721. whence x is not greater than 12. The string must form a semi-circle. (2) 24m = m' + (some number greater than 60). and Xl . Thus (1) 22m = m 1 + (some number less than 60). Then the entire closed curve will be the curve that encloses the largest area. pp. which is a square = (x . and x' . therefore 5z + 92 .60. with
. 'Let x = the whole number of measures. a fact which follows from the theorem that the polygon with a given number of sides. Now t of the price of the five-drachma measures + 1 of the price of the eight-drachma measures = x. say. Therefore the number of eight-drachma measures = 59/12.

. then the area surrounded by the four circular arcs cannot be a maximum since they no longer form a circle: yet the areas between the arcs and the sides of the quadrilateral have not changed .c) for the area of a triangle with sides a.
.. a..
.... is s.. and the figure moved. and this formula becomes Heron's formula ~)(s . then the area is given by
A = J(s ..only the interior area of the quadrilateral changes when the figure is moved about its hinges. band c. and the four arcs of the circle are then also hinged at the vertices of the quadrilateral. If half the sum of the sides.. and this will be so when it is a regular octagon. then the quadrilateral becomes a triangle. I
I
I I
"
I
'. The area enclosed by the screen will be a maximum when the area of the entire octagon is a maximum. The area can be calculated by a formula that was discovered by Brahmagupta but also apparently known to Archimedes. This conclusion is suggested by the thought that if the quadrilateral is adjusted so that its vertices do lie on a circle. to get this complete figure. and then reflect a third time.
maximum area.b)(s .. Reflect the original figure in both walls. which is certainly possible..
///
_-------" .. So the screen must be placed so that it meets the walls at two angles of 67}0 each..188
Penguin Book of Curious and Interesting Puzzles
\
\ \
. The area is a maximum when the ends of the rods lie on a circle.d)
(If one of the sides has zero length... c and d.. s is half the of the sides.. if the number of sides is then allowed to tend to infinity.a)(s . 31.....)
32. which is automatically inscribable in a circle.b)(s .c)(s .. b. is a regular polygon...

34. [Problems 34-6 are from The Greek Anthology. The father survived him for four years. 1941]
37. the husband 205.
38. the son 246. 577. Abul Wafa gave five different solutions. Then G is one of the other vertices. the husband receives five. Reflect the isosceles triangle in its third. variable. side. Construct N so that ABN is equilateral. and 422. If the estate remaining after payment of the legacy is divided into twenty parts. The stranger receives 300. Here are three of them. so Al-Khwarizmi divides the whole estate into 20 x 56 = 1120 parts. the son six and each daughter three. and draw an arc cutting ABF so that FN = FG.
35. 36. and each daughter 123.
5~
hours are past and
~
remain. He was a boy for fourteen years. The area of the rhombus will be a maximum when it is a square. Let one vertex of the equilateral triangle be at D. and so the area of the isosceles triangle is a maximum when the angle between its equal sides is a right-angle. dying at the age of 84. at 33 he married. and the last vertex is easily found on BC
.. Mark F on AB so that AB = BF. a youth for seven.The Solutions
189
33. and at 38 he had a son born to him who died at the age of 42. to form a rhombus. The stranger receives 15/56.

1986]
.
~----------------. The second bisector cuts AB at G. of DC. Bisect AD! and then bisect the half towards AD again. Let DN cut CB at H. Draw an arc with centre Band radius BA to cut BM at N. one of the other vertices sought.190
Penguin Book of Curious and Interesting Puzzles
F
Construct AID to be equilateral. Then H is one of the other vertices sought. M.
D~=-----~-------.
A~----------------~
[Berggren.c
~--~------~------~B
Join B to the mid-point.

then joining the dotted lines is one standard way of di~secting one small square and one large square into a single square. 40. Joining vertices by the dotted lines. the larger square is found.
This solution is not as idiosyncratic as it initially appears.The Solutions
191
39. 228 below) can also be seen as an example of Abul Wafa's theme. If one of the original squares plus a square composed of all four of the 'quarters' are repeated to form a tessellation. Note that the simplest dissection of a Greek Cross into a square (see p. Abul Wafa's solution bisects two of the squares and places them symmetrically around the third.
. Their size is irrelevant. The same solution works if the two larger squares are bisected. The four small pieces outside it fit exactly the spaces inside its boundary. It also works if the larger square is thought of as composed of four quarters.

with centres A and C. 1975] 42. and draw two arcs. Then AE is perpendicular to AB. Mark AC equal to the fixed radius. The dotted lines complete the dissection. Dissect the larger hexagon into six identical triangles. and join the vertices as indicated. the third vertex of equilateral triangle ACD.192
Penguin Book of Curious and Interesting Puzzles
41. 1975] 43. so that DE = CD. Arrange the three larger triangles round the small similar triangles. Extend the line CD and mark off E. as here.
.
[Wells. to construct D. like this.
[Wells.
and then arrange them around the smaller hexagon.

The Solutions
E
193
A
c
B
44. and so on. Joining the first mark on one perpendicular to the nth mark on the other.
. Construct perpendiculars in opposite directions at the endpoints A and B. will divide the segment AB into n + 1 equal parts. using the fixed radius. Mark off as many segments as necessary along each perpendicular.

924 Great Easterns. this is the same as the number of moves required to transfer all the rings in Lucas's 'Tower of Hanoi' puzzle (problem 238) and also roughly the number of ancestors that
. 0 is another.298 bushels of grain.411.500 tons) in the first week. that this is a total of 31. 1986]46. then bisect AD at Z and draw the line ZE.274.1 = 18. two in the next and so on. erect AE perpendicular to AD and on AE mark off AE = AD.073. Card Games and Fireside Fun (1881) calculates.744. perpendicular to EZ. the authors further calculate that if one pin were dropped into the hold of the Great Eastern steamship (22. As Kasner and Newman remark. and the perpendicular bisectors of MO and LO meet the circle at the other vertices. 'Finally. let the circle with centre I and radius AD meet the given circle at points M and L. a year's worth would fill 27. 'On this line mark off ZH = AD and bisect ZH at T. [Berggren.615 grains of wheat.551.'
4-4---~--~~------~O
M and L are two vertices of the required pentagon. taking an average number of 9216 wheat kernels to a pint. Then construct a line through T. 'a larger amount than the whole world would produce in several years'.709.446.194
Penguin Book of Curious and Interesting Puzzles
45. By way of a modern example. Cassell's Book of Indoor Amusements. 'At the endpoint of A of the radius DA. and let it meet DA extended at I. Sissa required 2'· .997.

six times in all.608. The trusts set the price to be (b + d)/(a + c). p. it follows that the trust price will be advantageous only if a > c and bla > dlc. [Mahavira. p. 1949] 47. this version is taken from Kraitchik. 35-6] 50. then the average price is Hbla + die). [The theme of this problem occurs in Mahavlra. 112] 49. pp. Similarly there are (6 x 5 x 4)/(3 x 2) = 20 ways of choosing three flavours. Comparing these two expressions. These are halved again and again. 1955. Then the problem states that
It follows w = 5. and c = 13. [Mahavira. and (6 x 5 x 4 x 3)/(4 x 3 x 2) = 15 choices of four flavours.
+w+c= that 5m + 2w
m
20
and
3m
+ !w + ic
= 20
= 20 and the unique solution IS m = 2.
48. and (6 x 5)/2 = 15 ways of choosing a pair of flavours.The Solutions
195
each person alive today had at the start of the Christian era. 150]
. The ratio of 2'4 to the actual population of the earth at that time is therefore a measure of the amount of unintentional interbreeding that has taken place. [Kasner and Newman. leaving 11128 = 1161. and simplifying. The total of all these answers. This last figure is equal to that for a choice of two flavours because choosing four flavours is the same as choosing two whIch you will not include. one half remain. 1912. By the same reasoning there are six ways of choosmg five flavours. There are six ways of choosing a single flavour. The total number of pearls is therefore the improbable 148. 1912. if the original prices are unequal and the denominator of the higher price is greater than that of the lower price. Fifty soldiers broke down and fifteen remained in the field. p. wand c respectively. including the single way in which all the flavours can be rejected is 2 x 2 x 2 x 2 x 2 x 2 = 64. because each flavour can be either rejected or accepted. that is. 1912. Let the numbers of men. [Mahavira. which happens to be about sixty-four generations ago. 73] 51. women and children be m. After: and! have reached the maid-servant and the bed. If bla and dlc are the original selling prices.

if the heights of the pillars are P and Q. q and r. then the required height is
1
or PQ
(P
+
Q)
(The point at which the string touches the ground divides the horizontal distance between the pillars in the ratio of their heights. and the value of the purse was 15. p.. of the series 1 + 6 + 12 + 18 + .196
Penguin Book of Curious and Interesting Puzzles
52.
53. and the wealth of the three merchants.. Mahavira does not state the distance between the pillars because this need not be known. From the equations
p
q
r
+ + +
x = 2q x = 3p
x = 5p
+ 2r + 3r + 5q
it follows that p: q: r = 1: 3: 5. The number of arrows in a bundle is the sum. and the solution in smallest integers is that the merchants originally had 1. [Mahavira.
If eighteen arrows are visible. Let the value of the purse be x. as far as necessary. 1912. p. that is. there are thirty-seven arrows in all. 167]
54.3 and 5 in money.)
. The height reqUired is one half of the harmonic mean of the given heights.

In general. [Midonick. 183. 16] 62. 18]
64. is a 12-13-5 triangle.198
Penguin Book of Curious and Interesting Puzzles
59. However. the only requirement of the problem is that the hypotenuse be 1 unit longer than another leg. the line from its base to the edge of the pool. 1964. respectively. p. [Midonick. 1964. p.21 measures of grain. or rather by the Gougu theorem. p. Mikami. p. Each ox. 183] 61.
4
~OOODoaooo/
9
5
1
2
7
I
+
o
I
3
8
6
[Needham.41. ¥f tae!' [Midonick. The solution is essentially unique. 1965. p. [Midonick. 1965. Twenty-three pheasants and twelve rabbits. but there are eight possible solutions obtained by merely reflecting and rotating anyone of them. which was the Chinese name for the theorem of Pythagoras. 57] 60. each sheep. 1* tael. Seven men and fifty-three articles. p. 1965. This is the original Lo Shu diagram. 183] 63. 1965. 184]
. 1959. By Pythagoras. p. The right-angled triangle formed by the length of the reed excluding the protruding foot. the formula (2n + 1)1 + (2n l + 2n)' = (2n ' + 2n + 1)' gives a right-angled triangle with hypotenuse and one leg differing by unity. the water is 12 feet deep. and the line from that same point on the edge to the point where the reed breaks the water.
[Mikami.
91.

184] 67. then x' +
Y
=
(10 .
199
4¥0 feet. respectively. p. by Pythagoras: the figure forms an 8-15-17 triangle.
10
x
3
[Midonick. 184] 66. demonstrated this solution by a dissection. [Midonick. Liu Hui. of dimensions a x b where a.
. If the height of the break is x. 1965. p. 1965. band c are the shorter sides and hypotenuse of the original triangle. The circle has diameter 6. 17 feet.The Solutions 65.
A
The triangle containing the inscribed circle is doubled to form a rectangle.x)'. the author of the Sea-Island Arithmetical Classic.

Out of this sequence. 18] 69. . i. 128.. 1964. pp. p.e. 233. Divide it by 13. p. also leave a remainder of 3 when divided by 5. [Mikami. 23.. 11. [Li Yan and Du Shiran. when we get 780. Sixty days. (where the difference is always 3 x 5 x 7 = 105) also leave a remainder of 2 when divided by 7. Therefore the smallest possible number of unknown things is 23.8. the lowest common multiple of 3. and length a + b + c. then I + J + i = 65. .. The numbers in this sequence all leave a remainder 2 when divided by 3: 2. . as Liu Hui proved by this beautiful figure. Here a = 8. all the numbers in the final sequence.26. The rule given by Sun Tsu is 'Arrange the 65 dishes. 1964. There were sixty guests. the sequence of numbers 8. p. the diameter of the required circle. Of these numbers. 1987. 17. 71] 68. 71. the side of the square is ab/(a + b).23.5. and multiply by 12. similar to the figure he used to solve problem 67:
.4 and 5. the numbers 23. 38. 31-2] 70. 33] 72. but there are in fact an infinite number of solutions to the original problem. [Mikami. 1964.. [Mikami.20.200
Penguin Book of Curious and Interesting Puzzles
The pieces are then rearranged to form a rectangle of height D. It follows that the required diameter is given by D = 2ab/(a + b + c). b = 15 and c = 17. days. when both grow to the same height of 4ll feet. . and thus we obtain the answer.. 2f... If the lengths of the shorter sides are a and b. 14.' This follows from the fact that if there were x guests.

However. He catches lip 37 . These equations are indeterminate . If the numbers of cocks. and some were sold of each kind. The Arabic author Abu Kamil. then the conditions are:
c
5
respec-
5c
+ h + 5 = 100 + 3h + 15 = 100
which together imply 7c + 4h = 100.
. 1987. miles. hens and chicks are c. written just before 900 AD."
I I
I. and this is then reassembled to form an equal rectangle of length a + b and height equal to the side of the required square. supposed that 100 birds are sold for 100 drachma.~____~__~~I ~ ________
cl_
L_____ J
~I
I
~I-
The original diagram is repeated to complete a rectangle a x b.The Solutions
A
201
---T-----. given that only whole numbers of birds were sold. so he will catch up 37 miles in 145 x 37/14 = 383t.____ -1 1
~----~---~---------~-----."1
1
I I
". p. in his Book of Arithmetical Rarities. 41] 74. then the number of possible solutions also increases dramatically..
1
I I
".. there are just three solutions:
c = 12 c = 8 c=4
h=4 h = 11 h = 18
5 = 5 5
84
= 81 = 78
If the number of kinds of articles sold is increased.' I ~. the birds being ducks at 2 drachma. but with the same information given. hand tively.1 1
1 I 1 1 1 1
I
1
1 . 70] 73.23 = 14 miles in 145 miles.'" .there are not enough conditions to fix the values exactly. [Li Yan and Du Shiran.. [Mikami. 1964. p..

1959.678.that 1 was reckoned a simpleton or an incompetent. At that time. Alcuin answers: 'First there were 250 pigs bought with 100 shillings at the above mentioned rate.
[Needham.. So 1 decided to write a book . for five fifties are 250. Chapter 12] 75. One sold the poorer quality pigs at three for a shilling. The actual number of solutions is 2. He wrote. ring-doves 3 for 1 drachma and larks 4 for 1 drachma. only to discover .' 77. and Abu Kamil is probably the first mathematician to have considered the number of solutions of a problem as a feature in itself. The one who sold the poorer pigs received 40 shillings for 120 pigs. However many footprints he makes in the earth as he goes forward. 'I went into [this problem] fully and found that there were 2.. This is Yang Hui's solution. 60] 76. doves 2 for a drachma. 1 marvelled at this.
. 1965. [O'Beirne. p.696 valid answers. and strangers looked upon me with suspicion. it was usually considered adequate to find one solution. 'The ox leaves no trace in the last furrow. each merchant had 125. Thus no footprint is revealed in the last furrow. the other the better quality at two for a shilling. following the example of Diophantos. On division. the cultivating plough destroys them all as it follows.202
Penguin Book of Curious and Interesting Puzzles
hens at 1 drachma. ' This must be one of the better reasons for writing a book.when 1 spoke of it . because he precedes the plough.

There then remained five of each sort of pig. leaving our sisters behind. This takes a total of eleven crossings. 80. 'First the two children get into the boat. the man whose sister remains on the first bank. Finally. 1 and my sister cross again and 1 return. the other women cross. and cross the river. and take the goat back over. 82. each having a parent who is sibling to a parent of the other. This is a total of nine crossings. and her child brings the boat back. give the first son the ten half-full flasks. the other two men cross. But divide them as follows. 'I would take the goat.8 . 79.. is: 1 cross with my sister. The other two sisters then cross. and return. and having left that behind 1 would take the cabbage across. . and so on. His brother joins him in the boat and they go across. The mother crosses in the boat. 4. there would be some healthy rowing.' 78. and my sister brings the boat back. and my sister returns. leave her on the other side.. 1 would then row again and having picked up the goat take it over once more. abbreviated.' 81.' 83. If the servant is not included in the count at each stage. in two ways. from which they could make a profit of 4 shillings and 2 pence. A shorter solution is: 1 and my sister cross. Then he and 1 cross over. Then the other two men cross and one returns with his own sister. 230 = 1. 1 and my sister cross over. 'To each son will come ten flasks as his portion.073. and picks up my sister who is carried over to us.
. and similarly to the third. then he would arrive at the first manor having collected no men. which is more than necessary. the total collected would be zero! Therefore the servant must include himself as the first soldier.. and on leaving the thirtieth manor. one of them brings the boat back. as the translator points out. By this procedure. and the numbers on leaving each manor are 2.The Solutions
203
the one who sold the better quality received 60 shillings for 120 pigs. so would collect none there. They are cousins twice over. my sister returns. crosses over and brings her to us.741. then to the second give five full and five empty flasks. Then 1 would return and take the wolf across . Alcuin's solution. and one of the women takes the boat back. and leave the wolf and the cabbage.824. and again one of them takes the boat back to his father.. 1 return. but without any lacerating catastrophe.

1(S13 . However. t and! of what they had first taken. The translator suggests adding the original fractions they expected. the average of t and is t so the mother receives 320 shillings. and this makes 100. There can be no integral solution. which is indeterminate. they possessed !S .
. and these amounts sum to S. Alcuin explains: 'Take the one which sits on the first step.. the mother receives 432 shillings. will always give 100 between the two. the sailing may be completed without shipwreck. and both cross again. tS . not having a pair. 87.x).' 84. Alcuin solves this much as Gauss solved. Since these are the sums that they possessed after putting back !. and find again 100. though it might be sufficient if we assume familiar legal principles not stated in the problem. The son and the daughter each receive one half of what they would have received if born alone. whose common difference is of the form 5p'.
n
n
85.x respectively. The information given is not sufficient. Fibonacci found 31'.204
Penguin Book of Curious and Interesting Puzzles
The father crosses. returns to his brother. and similarly the 100th is on its own. With such ingenious rowing. as is inevitable from the original conditions. the boy 336 shillings and the daughter 192 shillings. Before each man received a third of the sum returned. a rational solution can be found from any three squares in arithmetical progression. Also the second and the 98th. when a small boy in school. having boarded.x and 1S . which only concern proportions with no stated fixed amount. whose common difference is 720 = 5 X 12'. This gives the equation 7S = 47x.x). This problem is related to earlier problems (see problem 37 above) based on Islamic and Roman law.. Alcuin gives the mother the average of the two amounts she would have received in the two cases. 360 shillings and 280 shillings respectively. on division by p. and his son . The fiftieth step is on its own. Hence his solution is 41112. and add it to the 99 which are on the 99th step. the problem of quickly summing the integers from 1 to 100. i + + t (the mother's average expectation) for a total of f.x..x) and 1(S16 . that is. So for each step . Join altogether and get 5050.. 41' and 49'. Now multiplying their original expectations by i.' 86. Let S be the original sum and 3x the sum returned equally to the three men. the amounts first taken were 2(S12 .

The Solutions
205
Fibonacci chose the simplest values. are (382). 190. received (N . Suppose the last son received N bezants. The sums taken by the men from the original pile are then 33. 1976. [Eves.22. p. [Eves. This is equivalent to a cistern problem.1. supposing that the first pair bred immediately):
1
2 3 5
8
13
21
34 55
89
144 233
This is the famous Fibonacci sequence. working backwards. which was divided between six sons. 230]
91. so there were N + 6 bezants to be distributed after the previous son had received his share. who also received N bezants in total.4. 1981. 61 et seq. For many of its wealth of properties. and these N + 6 bezants make the two shares of N each taken by the last two sons. The last-but-one son. 231]
. There were therefore seven bezants remaining after he had taken N . 10. p. 1987] 90. so named by Lucas in 1877. Instead of three pipes pouring water into a pool at different rates. 166-7] 88. and 1 for himself.Js)" 2" x Js
89. p. three animals remove flesh from the sheep at different rates. [Eves. Therefore they will eat one sheep in 60/37 = 1~ hours. [Fauvel and Gray.1) + ~ of the remaining bezants at that stage. the father left an estate of thirty-six bezants. Fibonacci argues that in sixty hours (a conveniently chosen number) the lion would eat fifteen sheep.(1 . pp. He took 382 apples. Each term is the sum of the previous two terms. The numbers left after he gives one half and one apple more to successive guards. a total of thirty-seven. 1976. the leopard would eat twelve. Assuming that the rabbits are immortal. see the Penguin Dictionary of Curious and Interesting Numbers. Therefore N = 6 and. 13 and 1. the number of new pairs produced per month follows this sequence (Leonardo omitted the first term. Binet proved in 1844 that the nth Fibonacci number is given by the formula:
F = (1
+
"
Js)" . 94. and the bear ten. S = 47 and x = 7.46.

There is some uncertainty in this solution. of a day. So the total time taken is 63n. Draw a circle through the highest and lowest points of the statue. since the statue is not a vertical rod! Turning the figure on its side. days. The spectator should stand with his or her eye at that point. In 63 day-plus-nights it moves upwards 63 x (j -
t) = 147/5 =
29i. so that it touches a horizontal line through the eye of the spectator.
The final i will be covered in i -. 93.206
Penguin Book of Curious and Interesting Puzzles
92.
goal
line
T
. the solution for the rugby conversion problem is the same.. after which the serpent will not slip back.i = n. The conversion should be taken from T.

1976. the same angle will be subtended by the goal-posts at any point on this circle. Empty the 3 jar back into the cask and pour the 2 pints into the 3 jar. fill the 5 jar and fill the 3 jar from it. He worked for eighteen days and did not work on twelve days. p. Josephus and his companion placed themselves at positions 16 and 31 in the circle of 41 souls. leaving the middle cells empty. [Eves. 94. The Christians and Turks should be placed in the following circular order. Place two nuns in each of the corner cells. The speeds of the couriers are 250/7 and 250/9. 95.) Choosing the circle to just touch the line ensures that all other points of the line are outside the circle. so their speed of approach to each other is 250 (1/7 + 1/9) = 250 x 16/63 and they will meet in 63/16 = 3* days.
. a larger angle. in which the first person follows the last. which takes 1 pint.
2
2
2
2
96. 98. Next. Fill the 5 jar and fill the 3 jar from it. though not part of the usual statement of the theorem. any point outside it will subtend a smaller angle and any point inside it. By working backwards. and the counting starts with the first person: CCCCTTTTTCCTCCCTCT TCCTTTCTTTCCTT. (Both true. leaving 2 pints in the 5 jar. 99. 235] 97.The Solutions
207
Because of 'the angle in the same segment is equal' property. leaving 4 pints exactly in the 5 jar.

and the other pieces then chug round the circle in the same direction.
130
90
o
•
Full Empty
101. and the other eight squares can be reached from just two other squares.208
Penguin B. the central square can be reached by none of the knights.
From the circular figure. As Dudeney pointed out. in either direction.ook of Curious and Interesting Puzzles
o
100. It is trivially easy to read off the moves that must be made: any piece may make the first move. as illustrated by the circular figure.
.

[Gardner.3.b.2.e are: 7. It therefore made no difference whether the given fractions added up to less than one. take 112.3)/x. Let the barrel originally contain x pints of wine.5.3. 1978a) However.8. Lend 1 horse.3(x .9.2. and if you do not mind how many horses are borrowed or lent. For example. and the equation simplifies neatly to 2(x .3)/x)
x The total wine removed is one half the original quantity.only exists for certain conditions of the problem. Tartaglia gives a pheasant to Piero. share 77 horses by lending 17 and taking them back! 104.12. Tartaglia suggested the device of borrowing an extra horse.2. Previous authors had assumed that such conditions meant that the inheritance had to be divided literally in the proportions given. then the seven possible values of n. and then take your borrowed horse back. each person can now be given a whole number of horses.4. 41.a. 17. respectively. 113 and 115 of 30. 103. suppose you wish to share 31 horses in the ratios 112 to 113 to 115 (sum 31130).54 pints. to one exactly. If n horses are to be divided among three sons. and to Andrea's son Filippo.2. making eighteen horses. If the answer given to you is N. depending on whether you were told at the second step that the answer was even or odd.The Solutions
209
102. or to more than one. 19. then the number originally chosen is 2N or 2N + 1. 23. 11.2.3)/x.
105. who receive respectively lIa. lib and lie. its strength will be (x .1) = 14. For 112. 113. 11. if you do not mind whether you borrow-and-return or lend-and-recover. After one removal and replacement. and on the third removal.3. which is 15.2. totalling seventeen in all.4. A solution by this precise trick method . totalling 77/60.3. then the method always works whenever the numerator of the sum of the fractions is equal to the number of horses to be divided.7. the wine removed will be
3(x -
3 . and the borrowed horse returned to its owner. and the amount of wine removed on the second removal will actually only be 3(x .3)3 = xl.8. 10 and 6. and to Piero's son Andrea.
. 114 and 115.2.6. or x = (3 X 2'/3)/ (2'/3 .by borrowing exactly one horse and then returning it .4.

1. 32. then the original number is the remainder when 40A + 45B + 36C is divided by 60. multiplying the equations by 40.1 =3' + 3' . 1. This was proved in 1886 by Major MacMahon.1) . Therefore x and 40A + 45B + 36C have the same remainder when divided by 60.34 . in three stages: x and 3x. Suppose that the number chosen is x. Thus Bachet's solution does indeed use the fewest weights and is also the only solution in which all the weights are different. 8. 45 and 36 respectively:
40x = 120a 45x = 180b 36x = 180c
+ 40A + 45B + 36C
Adding. 1\ 94 . It is plausible that Bachet's solution is in some sense best possible.. if they can only be placed on one side of the balance.1.1 = 3(3 x 4 . The numbers of counters in the hands of the first and second person are. merely because it is so simple and elegant.9.. (x . it equals that remainder. The basic idea is that every number is one more or one less than a multiple of 3.3 . 1\ 9. so that the three equations are x = 3a + A = 4b + B = 4c + C. This is the same as saying that each integer can be represented uniquely in the binary notation.27.94 . Bachet gave the solution 1. are A. 32. 121x = a multiple of 60 + (40A + 45B + 36C). Denoting the number of each weight by a superscript.5) + 3(x . 4. 108. and since x was chosen to be less than 60.4 . 1. Band C respectively. Twenty counters. Tartaglia (and Fibonacci before him) had considered the problem of the weights required. 4 and 5. who used the method of generating functions discovered by Euler to show that there are eight possible sets of weights. Thus 32 = (3 XII) . x .1) .5).3(x . P\ 27. 107. and the last number is always 20. If the remainders on dividing by 3.27. and so on.1 = 3(3 x (3 + 1) . 1.3.
. 1. Therefore 32 Ibs can be weighed by placing the 27 and 9 Ib weights in one pan and the 3 and lIb weights in the other.3. 27. 16. and concluded that the best solution has weights in the sequence of powers of 2: 1. 9 and 27 when both pans may be used.210
Penguin Book of Curious and Interesting Puzzles
106. they are: 140 . 3.313. apart from the one-scale solution.5 and 3x +5. 2. Then.5) and 3x + 5 .2.

111.
110. The obtuse-angled triangle with sides 4-13-15 has area 54 . q = 2mn and r = m ' + nl.
(4A)'
=
. The right-angled triangles 5-12-13 and 6-8-10 each have area equal to perimeter. The easiest way to construct Heronian triangles is to fit two right-angled triangles together.The Solutions
211
109.30 = 24. In the solution to 109. place the same two triangles so that they overlap.nl. 7-15-20. The area is 84.
5
9
Here are a 5-12-13 and 9-12-15 triangle fitted together. The resulting triangle has altitude and sides 12-13-14-15. p m' . 9-10-17. The three proper Heronian triangles with this property are 6-25-29. and is the only possible such triangle.a ' )' = (2bc)' This is of the form p' + q' = r and has parametric solutions. Proofs of these results are not so simple. one method is to write Heron's formula in the form
which can be written in the form
+ (b ' + c' .

So arranged. The centre of gravity of the entire arrangement will then be below the finger tip and will be stable.
115. which can then be lifted. so that the knives hang well below the finger. Bend the straw and insert into the neck of the bottle. they can easily support a glass of water well above the table surface. 114.
113.
. Each knife blade goes under the blade of one other knife and over the blade of the third.212
Penguin Book of Curious and Interesting Puzzles
112. Force the tips of three knives into the stick.

in which the combination of holes to be plugged are different.
117. He is standing at the centre of the earth.
. The principle is the same. circular and oval.
Van Etten gives variants of this puzzle.The Solutions
213
116. such as square.

One horse travelled east and the other travelled west. then one hole must contain at least two objects. 125. This construction works. which will be smaller than the circle drawn by the same compass on a plane surface. Choos-
. place the point of the compass at the apex of a circular cone. they would also be moving apart. (This is a general principle that applies to all such minimum networks of 'roads'. The compass will then draw an oval. The sum of the distances will be a minimum when the lines OA. 124. the point sought is at that vertex. you will be forced to pick an individual with a number of hairs that you have already counted once. which is O. Wrap the paper on which you are to draw the oval round a cylinder. 121. 123. which says that if you have N + 1 objects to place in N pigeon holes. and draw the circumcircles of each new triangle. the first gaining in the number of days it lived. draw equilateral triangles outwards on each side of the triangle. then later in the day they sold them at 3p each. and draw a circle on its surface. and C sold 32 @ 1 p and 8 @ 3p. B sold 17 @ 1 P and 13 @ 3 p. and the second losing. First they sold their apples at 1 penny each. each also making 56 pence. because of the property that the opposite angles of a cyclic quadrilateral sum to 180°. and ascending in opposite directions. 08 and OC all meet at 120°. 120.214
Penguin Book of Curious and Interesting Puzzles
118. it is also true that if they ascended two vertical ladders on the earth's surface. A sold 2 @ Ip and 18 @ 3p. Therefore if you start to pick out individuals with given numbers of hairs on their heads.they are at the centre of the earth.) To construct point 0 with ruler and compasses. The number of individuals in the world far exceeds the number of hairs on the head of anyone of them. 119. Place the point on the surface of a sphere and draw a circle. long before the population of the world is exhausted. When he is standing at the North Pole. Van Etten gives the same solution . making 56 peace. This is the first known example of the 'pigeonhole principle'. These circles will pass through a common point. albeit by a minuscule amount. If one of the angles is greater than or equal to 120°. However. 122. Alternatively.

1973. p.The Solutions
215
ing the extra triangles to be equilateral. 24] 126. with angles of 60°. It is only necessary to 'tie' an ordinary knot in the strip and carefully flatten it. they will all pass through the point O. ensures that the opposite angles will all be 120° as desired.
. It also happens that if the outer vertices of the equilateral triangle are joined to the opposite vertices of the original triangle. [For further discussion see Honsberger.

. . .. Then and and
bM b'M b"M
+ + +
cbm ...060660.. Newton expresses the problem in general form.. 128. can be passed through a given cube of side 1 unit.. M. rather than giving values to the letters. ~ . and that each cow consumes the same amount of grass per day. m and q are not all zero.c"a"Q = 0
By a standard theorem in determinants. the bottom face along ABCD and the two vertical edges at X and Y. the square hole cuts the top face along the lines EFGH.c'a'Q = 0 c"b"m . Suppose that each field contains initially the same amount of grass. and the daily growth in each field is also the same. given that M. m.216
Penguin Book of Curious and Interesting Puzzles
127... Amazingly. this determinant is zero:
b b' b" bc b'c' b"c" ca c'a' c"a"
... •
In this figure. . Typically.
...caQ = 0 c'b'm .... as indicated by the dotted lines... the answer is that a cube of side slightly under
3j2
4 or approximately 1.... Q. """..

'The probability of 1 or more sixes with 6 dice is the complement of the probability of 0 sixes: 1-
(~)G)O(D6 ~ 0. Fortunately. p.3. 1965.4. Newton had to work the probabilities out by hand.e')
= 0
[After Dorrie. This is not so.The Solutions
217
Without usmg determinants. m and Q can be eliminated 'by hand'.5. . because the proportion of sixes required to the number of dice thrown is constant.1. It might seem that the chances are equal. Harvard University Press. Our
.ba')
+ e"b"(be'a' . and one of the tabled values is p = i.665
1_
'When 6n dice are rolled.1.b'ea) + e"a"bb'(e . Therefore the probability of exactly 1 six is
mW
(~)G)(~Y
Similarly. n
This formula gives the terms of what is called a binomial distribution.. to give
b"ee'(ab' . . 1955.6
The probability of x sixes for n dice is
x = 0. this table gives the cumulative binomial for various values of p (the probability of success on a single trial).2. 9] 129. the probability of exactly x sixes when 6 dice are thrown is
x = 0. the probability of n or more sixes is
I
x="
(6n)(~)X(~)6"-X
x
6 6
'i (6n)(~)X(~)6"-X
l
<=0
X
6
6
Unfortunately. M. 'The chance of getting 1 six and 5 other outcomes in a particular 5 • We need to multiply by the number of orders for 1 six order is and 5 non-sixes. but we can use the Tables of the Cumulative Binomial Distribution..

3! + 4! .5! + .665 0.
6n n P{n or more sixes)
6 12 18 24 30 96 600 900
1 2 3 4 5 6 100 150
0. he decided to welch on his original bet.. p. 22]
132.597 0.542 0. When he found that out. of obtaining the mean number or more sixes when 6n dice are tossed.619 0. the value is 1854. but the principle is the same.576 0.' [Mosteller.
[Fisher. [A complete solution is in Dorrie. 1982. p. These two solutions have the features of simplicity and symmetry.218
Penguin Book of Curious and Interesting Puzzles
short table shows the probabilities.517 0. + --.. rounded to three decimals. 19]
131.-
1
1
1
1
-1")
When n = 7. 1987. . The general formula for the number or ways of misplacing all the letters is:
n! (
2i . 1973]
.584 0. 1965. [Newman. One! This answer is independent of the number of letters and envelopes. problem 19] 130. Bernoulli posed the problem in terms on n letters wrongly placed into n envelopes. .514
Clearly Pepys will do better with the 6-dice wager than with 12 or 18.

Therefore. any vertex at which an odd number of edges meet (all of which must be traversed) can only be a starting or an ending vertex. This is one solution. The map of the river can be represented schematically like this:
The problem is now to trace out this figure with a pencil. But the number of edge-ends is also the total of the number of edges meeting at each of the individual vertices. This was one of the points established by Euler in his original paper. of which there can be at most two. and trace the figure so as to return to your starting point). No. The reasoning IS simple: in arriving at a vertex and then leaving it. in which case you can only trace the figure by starting at one and ending at the other. and no line twice.The Solutions
219
133. This is only possible if the figure to be traced contains either no vertices at which an odd number of edges meet (in which case you may start at any point you choose. which must therefore include an even number of odd vertices. All four vertices in the figure for the Bridges of Konigsberg are 'odd'. and the patterns formed by the shading are similar. Every edge has two ends. 134. it is not. two of the edges meeting at the vertex are 'used up'. or just two such vertices. without lifting your pencil from the paper. and so the figure cannot be traversed.
. since an odd number of odd vertices would give an odd grand total. Typically the same letters are knight's moves apart from each other. passing over every line once. 135. so the total number of edge-ends IS even.

But the latter is the product of three factors whose sum is constant. P. call AP x. As the editor remarks of Question 97. find a point on it. (Readers also occasionally stooped to taking questions from published sources. such that AP' x PQ is a maximum.x).
138.2x and x = 2L13. For clarity.220
Penguin Book of Curious and Interesting Puzzles
A
C
E
B 0
137. and it will achieve its maximum value if all three factors are equal. The given solution is something of a cheat. at 2L. it is required to maximize AP' x PQ.
B C 0 E D E A B A B C 0 C D E A E A B C
136. So AP must be two-thirds of AQ. given any line AQ.013 years. Sixty-six years.
. quickly referring to one of Dr Hutton's textbooks. Forty-five and fifteen years. 144 days. 5 hours and 55 minutes. In other words. as a proportion. 'It is evident that this question is composed from that in page 225 of Ward's MathematicIan's Guide'!) Assuming that the cylinder to be cut out has its base on the base of the cone. Therefore. and let the height AQ = L. 139. Then we have to maximize xZ(L . Therefore the maximum is when x = 2L . which is the same as maximizing xZ(2L . The volume of the cylinder is proportional to DF x PQ. and we know that DElAP is constant. it remains only to determine its height.2x). 19.

which are 5/13 and (8-5)/8 = 3/8.minutes past 7 o'clock. By taking four consecutive terms later in the sequence. as exaggerated here.
.9. the numbers of hogs bought by husband and wife are such that the differences of their squares are 63. the paradox becomes even more difficult to detect by eyesight alone. noted. 143. which gives three possible pairs only: 8. These fractions are close.1. 12. or approximately 7. the products of the outer pair and of the inner pair differ by one. but not equal.
8
5
5
5
5
8
This can be checked by calculating the slopes of the different portions of this 'diagonal'. 32. The extra square has actually become the long and very thin parallelogram formed by the 'diagonals' of the second figure.31. which has the property that of any four consecutive terms. Farrer.222
Penguin Book of Curious and Interesting Puzzles
As another solver.. Mr N. The fractions are so close because they have been conveniently taken from the Fibonacci sequence. 1 1 2 3 5 8 13 21 34 55 . 142..05 and 27 seconds. 5rl:.

their meeting would be at the mid-point.8. 147.
149. One-third of TWELVE is LV = 55 in Roman numerals. 150.
. downstream from Haverhill towards Newburyport. 123456789 x 8
=
987654312. 1978) 145. onefifth of SEVEN is V = 5.The Solutions
144. and will close the 18 miles between them in 21 hours. From
99~. 2
+4+6+
0. so they approach each other at 8 miles per hour. The rowers move at 4 miles per hour relative to the water. their meeting place will have moved at H miles per hour. If the water were still.
223
[Delft and Bottermans. But due to the current.
SIX IX XL
take
IX X L
leaving
S I X
148. So they meet 5i miles from Newburyport. over the 21 hours. 146. and 55/5 = 11.8 = 12. a total drift of 3~ miles.8 and 1
+
3
+
7
+
9/5 = 12. 9 miles from each town.

4. This is jackson's solution. 1. one solution. 8. So. If each number is replaced by the matching power of 2. 8. then it is magic by multiplication of rows and columns. top to bottom) 7. 4. This is jackson's solution:
4
9 6
7
5
16
3
2
15 14
1
10
11
12
8
13
. arranged in tabular form:
barrel first container second container
12 7 7 2 2 9 9 4 4 11 11 6 6
5
5 3 3 5
5 5 7 3 3 7
5
1 6
152. 1. 5. 153. If cells are filled with the numbers (reading left to right. top to bottom. 2. 3. 6. 16.32. is: 128. then the square is magic in the usual way. 256. 2. 64. 0. reading the rows from left to right.224
Penguin Book of Curious and Interesting Puzzles
151. by addition of the rows and columns.

158. It generalizes to dividing a circle into N parts by N lines of equal length. 155. and the true weight is W. the square root of 16 x 9 or 121hs. X. Suppose that the long and short arms of the balance are of length p and q respectively. that is. A shoe. Divide a diameter into N equal parts.
Then DO is the length of the side of the required square.The Solutions
225
154. mark off the points C. 0 is the centre of the circle. This is jackson's solution. following this pattern.
.
157.2 snakelike regions between will all have equal area. 156. and similarly marking F. produces the square. Then with radius BC. Marking AE so that AE = DO. one of the poles and any two points on the equator which are separated by one qu~rter of the earth's circumference (taking the earth to be spherical). and construct a sequence of semi-circles on either side. for example. Then Wp = 16q and Wq = 9p. and will be bounded pairs of lines each equal in length to half the perimeter of the circle. Jackson merely states that the true weight is the mean proportional (or geometric mean). from which W'pq = (16 x 9)pq and the conclusion follows. and B in succession. The tadpole-shaped regions at each end and the N . draw arcs to intersect at
D. and centres A and B. With the compass open to the radius of the circle. taking. Draw it on a sphere.

161. quite arbitrary. a second 10 degrees on this side. since AOE = 90° and AOC side of an inscribed dodecagon.' The figures chosen by Jackson are. 162. of course. from Dover to Calais. since the pole contains all degrees of longitude. worthy of Sam Loyd. The fact is that any place on earth.226
Penguin Book of Curious and Interesting Puzzles
D
A
B
F
Also. (Any of the other Channel Islands would do. the poles apart. 160. based on the earth having a radius of about 4000 miles. COE = 30° and CE is one
159. 3000 miles is jackson's estimate of the variation for Naples. and a third 20 degrees on the other.
= 60°. but England. The South Pole. and likewise in longitude.) Guernsey is 26 miles from France. varies daily in distance from the sun because of the earth's rotation. under the same meridian circle. The island is Guernsey. The reference to Naples and the situation of the village in a low valley are mere Rim-Ram. being a maximum for places on the equator. 'Suppose one place to lie directly under either of the poles. then they will all differ in latitude. is only 21 miles from France. and a minimum of zero for the actual poles.
.

. The traveller's journey has been right round the world. by their own reckoning. and so the radius of the giant circle travelled by his head is about 6 feet greater than that of the circle travelled by his feet. having respectively lost a day and gained a day while travelling round the world. and arrangements of tiles could be turned over. 165.)
l/1SJ21 NZ1SJ
rsrsN l/1/1/1
tsN/1 l/1/lSJ
~~~6¥ ~~~6¥
. they will each be able to celebrate their own sabbath on the same day in the same place! 164. 2A-.. travelling East. but travels West.. 3rl. At In. or about 36 feet = 12 yards. hours. There are fourteen different arrangements. (Some pairs of these arrangements would be equivalent if the tiles were painted on both sides. ignoring colours. each of which can be coloured in two ways. 166. The Christian sets off from the Jew's abode. and the Turk does likewise. His head is about 6 feet from the ground. Jackson attributes this idea to Whiston's commentary to his edition of Euclid. This difference in radius produces a difference in circumferences of about 27t x 6 feet. When they meet again at the Jew's. making a total of twentyeight.The Solutions
227
163.

This solution is equivalent to the problem of dissecting a Greek Cross into a square: assemble the five squares to form the cross.-
167. there are an infinite number of solutions.
This dissection of a Greek Cross is one of an infinite number which depend on the possibility of tessellating the plane with identical crosses:
Take any four corresponding pomts to form the vertices of a square. A: 14¥.228
Penguin Book of Curious and Interesting Puzzles
B: 17* C: 23f. equal in area to any Cross. and the solution appears immediately. and the square at once forms a dissection.
168. Since the four corresponding points can be chosen in an infinite number of ways.
.

8) of the second. when
~ R' = R n
2
and
R =
3
~. that is. Imagine that equal weights are placed at the vertices of the original polygon. 175. Then the triangles PXP' and QXQ' will have equal areas.
Therefore the surface area is n(R 2
+
~ + ~). and its top edge joins the mid-points of the other two sides. The largest possible rectangle has its base on one side. will not change the centre of gravity of the arrangement. which is fixed.
. that is. and the sequence of polygons contracts to a point that is the centre of gravity of equal weights at the original vertices of the polygon. so that PX = XQ. is V = nR2H. and the volume. or 8s a bottle.5) of the first to 2( = 10 .230
Penguin Book of Curious and Interesting Puzzles
171. The position PXQ is therefore a limiting point for the area of the enclosed triangle. Imagine that the line is rotated very slightly about X to cut the two lines at P' and Q'. to a first approximation. 3 bottles of the first costing 30s and 2 bottles of the second. costing lOs. Then the total surface area is nR 2 + 2nRH. Its area is then one half of the area of the triangle. This intuitive answer is confirmed by calculation.
172. Then replacing these equal weights by an identical set. X. The bottles must be mixed in the proportion of 3( = 8 . Let the radius of the base be R and the height H. 173.
I~n. Therefore the centre of gravity is unchanged by any number of repetitions of the process. 174. and since it cannot be a maximum.
nR nR
This is the sum of three terms whose product is constant. Draw a line through the given point. 5 bottles costing 40s. it must be a minimum. and therefore it makes no difference which is chosen to be the base of the rectangle: three different maximum-area rectangles have equal areas. It will therefore be a minimum if all three terms are equal. total. placed at the mid-points of the sides.

89. but only alternate games! 178. 1711 179. which. 23. 46152837.232
Penguin Book of Curious and Interesting Puzzles
177. and one on a 6 x 6: 246135. then of course you must lose. there is one basIC solution on a 4 x 4 board: 3142. the twelve solutions are: 41582736. 42751863. The secret is to make the sum. however. Each can be described by a single 8-digit number. 45. 48157263.
. 12. Once you have achieved this (which is easy enough if you start the game you just choose 1 as your first call). If your opponent starts the game. On a 7 x 7 board there are six basic solutions. by reading off the position of each queen in each column. then you can keep to the sequence by calling out the difference between 11 and your opponent's last call. will in some cases be identical to the original solution due to its symmetries. 42736851. 42861357. 41586372. 47526138. equal to a number in the sequence 1. There are twO basic solutions on a 5 x 5: 14253 and 25314. 78. Using the same notation as solution 178. starting from one end . after one of your turns. With this notation. and knows the trick. Every solution can be rotated and reflected to reproduce seven other solutions. 42857136. 34. 42736815. 46827135. 100. [Rouse Ball. 42586137. 67. p. 1974.
There are twelve basic solutions on the full chessboard . 56.

then the points P.
p
The pairs of corresponding sides will then also meet in pairs on a line L. which says that if points A. and A'.) If it happens that B. B' and C' taken on another. such that the lines through matching vertices meet in a point P. (P is the meet of BC' and B'C. This is the commonest solution offered:
233
It is a special case of Pappus's theorem. Q and R will lie on a line. In other words. and joined as in the figure. then the figure satisfies the conditions of the problem. Q and B' also lie on a line. A more general solution is any figure for Desargues's theorem. take any two triangles. and so on.
. B and C are taken on one line.The Solutions
180.

and 3 on 1. any two overlapping triangles will do.
Penguin Book of Curious and Interesting Puzzles
* * * * * * * * *
* * *
'*
* * * * * * *
0
* * *
*
182. Indeed. 3 on 7. at equal distance from each other. 185. Plant three of the trees at the base of a steep mound. The same trick may be thus performed: place 5 on 2. 12. at equal distance from the other three. 6 on 2.
184. 8 on 6. 'Place 4 on 7. The four parts are 8. and plant the fourth tree on top. 5 on 8.234 181. This solution (ignoring the dotted lines) can be varied. 183.'
. by varying the triangles. and 4 on 1. Recollect always to begin with either 4 or 5. 5 and 20.

The Solutions

235

l~·N

N E HUT
VII VIII

187.

-~---j---I---

-><------1-+--

188. This puzzle was the basis for one later made famous by Sam Loyd.

236
189.

Penguin Book of Curious and Interesting Puzzles

190.

-

I
r--

I

191. This problem has reappeared many times, often in its simplest guise - an L-shape formed from three-quarters of a square, and the demand that it be dissected into four identical parts.

The Solutions

237

192. Fold one edge of the square on to its opposite edge to get a middle line. With the middle line horizontal, fold one lower corner on to the middle line, so that the fold passes through the other lower corner. Repeat, using the other lower corner. These two folds and one edge form an equilateral triangle. [Tom Tit', n.d.] 193. 'Roll the paper into a short compact roll. Make two parallel cuts across the roll, each being about one half an inch from the other end. Then make a long cut parallel to the axis of the roll and terminated by the cross cuts. This will produce a gap in the roll. Holding the roll lightly in the fingers ease out the ends of the first strip, which lies at the bottom of the gap, then, taking the strip with the teeth and holding the roll lightly by its two ends, slowly draw the strip out of the gap ... the whole inside of the roll will be drawn through the gap, the connecting parts of the successive strips being twisted. The final result will be a series of paper strips which serve as rungs of a ladder, whose upright sides are formed by the twisted parts.'

194. Trace round a shilling and cut out the circle to make a circular hole. This hole cannot be increased in size as long as the paper is Rat, without tearing the paper. However, if the paper is folded across a

238

Penguin Book of Curious and Interesting Puzzles

diameter of the circular hole, and a larger coin - the half-crown was just the right size - is placed within the fold next to the hole, then by bending the paper without tearing, the hole can be enlarged sufficiently to allow the larger coin through. The maximum diameter of the hole when the paper is bent is fl, or very slightly over 1.4, times its original diameter. 195. Draw a rectangle inside the given sheet so that there sufficient margin for gluing the resulting envelope.
IS

a

Bisect the shorter sides, and draw a circle, centre at the centre of the rectangle, passing through A and B, to cut the other sides at a pair of opposite points, C and D. Join ABCD. This is the front of the envelope, which is just covered completely at the back, when the four outer triangles are folded inwards. 196. Like this (the principle can be used to cut any board in the same or related proportions):

2

4

I
4

2

The Solutions
197.

239

0 0 0

0 0 0 0 0 0

0 0 0

198. Fold the card in half, and make cuts with scissors as shown. Finally cut down the original central fold, omitting the two end portions. The card is now reduced to a strip which may be opened out and passed over a person.

199. Number the coins from 1 to 10, in sequence. Place 4 on 1,7 on 3,5 on 9, 2 on 6, and then 8 on 10.

203. Four strokes. 204. Scratch the table cloth and the 20p coin will emerge from under the edge of the tumbler. 205.

206.

ONE LOVE

The Solutions

241

207. Before picking up the handkerchief, fold your arms. Then pick up the handkerchief, and unfold your arms. The 'knot' which was in the folding of your arms is transferred, as it were, to the handkerchief. 208. Fold the piece of paper on which the sum is written, like this, so as to obscure the figure 300, and so that the other two lines form the new number 707. The new sum has the same total as the old.

3/8 701

209. Jump 9 over 4, 5, 7, and 1; 3 over 2; 6 over 8 and 3; and 9 over 6. 210. Arrange the matches to form the edges of a regular tetrahedron.
211.

o o o o o o o o 0

. respectively. which on being divided horizontally along the middle line become a row of zeros. Bet. The number of letters contained in each numeral word. 2s 7d. 15 11d.242
212. 216. 215. Jane. 1s 1d and 8d per week. Ann. or nothing. The four figures are 8888. 15 5d. (c) Draw it round his body. 3s 2d. Rose and Jim earn. (a) 19!
(b) Place one of the coins on the table.
Penguin Book of Curious and Interesting Puzzles
213. then keeping the hands apart. take it up with the other hand. 217. Joe. (d) 8!. 214. The squirrel takes out each day one ear of corn and his own two ears.

plus such trade profit.
221. 142857. the nett value of the hat.
243
219.i85714. This period has the property that any circular rearrangement of the digits is a multiple of the period of t.
. The reply of most people is. 35170
+
148/296
=!+t =
1
222. that the hatter lost £3 19s Od and the value of the hat. The difference in their ages. 223. which is the period of the decimal fraction t = 0. His actual loss was £3 19s Od less his trade profit on the hat. The answer is 285714.The Solutions
218. The original puzzle states that the two digits transferred are 28. but a little consideration will show that this is incorrect. almost invariably.

the old gentleman thus became his father's brotherin-law.'~ . In the original figure there are a total of thirteen triangles. Remove the dotted matches. and the old gentleman thus became his father-in-Iaw's brotherin-law.
•
225.' 224.~ ~ ~
. 21s. three medium-sized and the outer triangle are left.
~ . The old gentleman and his father (who was also a widower) married two sisters (the wife of the old gentleman havmg a daughter by a former husband). thus the old gentleman became his brother's father-inlaw. The old gentleman's brother-in-law married the old gentle-
l5l~
/\
. The old gentleman's brother married the old gentleman's step-daughter. The old gentleman's father-in-law married the old gentleman'S sister.'~ .
L\\~6".. a total of seven triangles remaining./"'~
226.
. and three small triangles. The old gentleman was a Widower with a daughter and sister.
c=====~.244
Penguin Book of Curious and Interesting Puzzles
being balanced by the difference. which he retained out of the proceeds of the note.

First. The '*' symbol indicates that a piece has jumped. De Morgan was born in 1806. the front half of train B backs into the cul-de-sac. Third. 230.
228. and moves well forward of the junction. Interchange C and D by moving round the shaded cells and then shunt B-A to the left. uncouples it. train B advances. Move B one square to the right and move A round the circuit to the right of B. which is easily read off the plane map. move to the 'star' pentagon and go round it. There is essentially only one solution. The maximum value which cannot be made is 63. 64 = 2 @ 17 + 6 @ 5. the distance between each dog and the dog he is chasing will be reduced from the initial 100 yards at the rate of 3 yards per second. leaving the central cell empty: Hhg*Ffc"CBHh*GDFfehbag"GABHEFfdg"Hhbc"CFf"GHh* 233. A passes the junction. Yes. Second. Train B can now leave the cul-de-sac and join its rear half and proceed. and joins to the rear half of B. 235. and by symmetry they will collide at the centre of the field. 231. and so was 43 in the year 1849 4Y. Note that it is generally only necessary to name the piece that moves at each turn. due to Dudeney. They will therefore meet in 33! seconds. and then end by going round the outer pentagon. 864 . Higher numbers can always be made. 234.72
=
792. thus. This is a minimum solution. when the symmetries of the dodecahedron are taken into account. In forty-six moves. One person received his herring on the dish. backs. and backs its rear half into the cul-desac.
=
232. train A uncouples the rear half of train B and proceeds on its way. whereby the old gentleman became his brother-inlaw's father-in-law. Fourth. 227. 229. which it then draws out of the cul-de-sac and backs to the left. go round the minor pentagon.The Solutions
245
man's daughter. However the dogs run. For example.
.

provided that their length is at least
2J2W
3
where W is the width of the moat (with a little allowance for the planks to overlap each other and the bank).
237."'::""::"4":::". No. The planks can be used together.246
Penguin Book of Curious and Interesting Puzzles
236. The solution can be represented visually.
x
w
ff
Since 8 <
2J2 x 10
3
the given planks will not suffice. like this:
New York~~::::"-:-::~':::""~::.:~::"':::'''''::::''"""T-r­
o
1234567891011121314151617 days
.

3-C. 2. therefore. problem 255] 238. The principles are the same whatever the number of discs.000 years. and an even numbered disc. It passes thirteen ships at sea. Following this rule. a total of fifteen. Number the discs 1 to 8 from the top downwards. from peg A to peg B. 4-B. I-B (leaving the first four discs on peg B). Suppose. I-A. which at one move every second amounts to more than 500.. In Lucas's original story the number of moves required is therefore 264 . Always move an odd numbered disc. Let Band W' stand for the black or white counter which is in the bag at the start and W' for the added white counter. and similarly for the remaining discs. It is therefore 2" . on its {irst move from peg A. The probability is i.
1. disc 2 visits them in the order B-C-A-B-C-A so that its visits 'rotate' in the opposite direction. . place disc 7 on C and repeat to get all discs up to 7 on C. disc 3 every eighth move. Then place disc 6 on Band repeat to get all discs up to 6 on B. 239. to leave all discs up to 5 on C. 2-B. I-B. The next move is to place disc 5 on to peg C and repeat the process so far. 1972. and finally put disc 8 on peg B and repeat the entire process to transfer the seven smaller discs from peg C to peg B. on its {irst move from peg A. one of Gardner's readers.1. solving the problem for just two discs). the sequence for eight discs will start: l-C. I-C. for simplicity. 2-B. The simplest rules are: . The following explanation is due to Howard Ellis. to peg B. [Attributed to Lucas in Kordemsky. one in each harbour.. and meets two more ships. that there are eight discs to be moved. 2-A. I-B (the first two discs have now been transferred to peg B. The number of moves taken to move n + 1 discs is one more than twice the number needed to move n discs. I-A. disc 2 every fourth move.The Solutions
247
The line AB represents the ship leaving Le Havre today. 2-C. l-C (leaving discs 1 to 3 on C).1. Move disc 1 every second move.000. 3-B. Notice that disc 1 visits the pegs repeatedly in the order C-B-A-CB-A .
to peg C. and so on.000. After removing a white counter there are three equally likely states:
.

1981. the chance of theIr lying in a straight line being (practically) nil. 1958. This is equivalent to asking for the relative volumes of a regular tetrahedron and a regular octahedron. p. Carroll solved this problem by calculation. and the octahedron occupies the other half. [Carroll. 189] 240. [Carroll. 1958. and Gardner.
The complete tetrahedron has 2 x 2 x 2 = 8 times the volume of any of the four small tetrahedra affixed to alternate faces of the o'ctahedron. The pyramid which is one half of the octahedron therefore occupies one quarter of the complete tetrahedron and is equal in volume to two of the small tetrahedra.
. This is Carroll's own solution: 'It may be assumed that the three points form a triangle. a white counter remains in the bag. problem 5. problem 49] 241. Fit two such pyramids together to make a regular octahedron and inscribe it in a regular tetrahedron. Those four tetrahedra therefore occupy in total one half of the volume of the large tetrahedron.248
Penguin Book of Curious and Interesting Puzzles
In bag Outside bag
WI
W'
WI
W'
B
W'
In two out of three cases. but it is solved more efficiently by visualization.

'Also. problem 58] 242. The clock that is losing time is correct once every two years.. chance
n/2
3 8
~n
3
.. the triangle is obtuse-angled. acute-angled.The Solutions
249
'Take the longest side of the triangle and call it AB.'. three of each of the 100
. Adding the four percentages together.. ABDCE 'Now let AB = 2a: then area of semicIrcle = Figure ABDCE = 2 x sector ABDC .area of Fig.. AEC. 1961..triangle ABC. 1958. Then it is evident that the vertex of the triangle cannot fall outside the Figure ABDCE. if it falls inside the semicircle.
=
It.. and
area of
2(4~') _ J3a1 = a{:n . Distributing them as evenly as possible.. intersecting at C.J3}
= ----
. whereas the stopped clock is right twice a day. (The chance of its falling on the semicircle is practically nil. the total is 310 per cent. BA.J3
6J3·
n
[Carroll. with centres A. Ten. B. and on that side of it on which the triangle lies draw the semicircle AFB. every time that 'its' time comes round! [Carroll.) area of semicircle 'Hence required chance = . 6] 243. p. Also. draw the arcs BDC. if outside it. and distances AB..

1961. [Carroll. [Carroll. 268) 247. or indeed not at all. Assume. p. Provided friction is neglected. and the daughter gets into the same basket as the son. the son descends against it. p.250
Penguin Book of Curious and Interesting Puzzles
per cent total. 1961.
~
minutes. p. there remains at least 10 per cent with all four disabilities. the weight descends. Finally. The daughter steps out at the top. and the daughter descends against son and weight. [CuthweIlis. the son remains and descends against the weight. as is necessary but also implied. Raise one basket. p. raising the son and the weight. 1978. that when one basket is within reach of the window. raising the second basket. [Cuthwellis. The son steps into this basket at ground level and the daughter descends in the other basket. The son gets in. 9)
245. into which the son steps and descends to the ground. 318)
. the son gets out. Lower the weight again. The answer is always £12 18s lId. the other is at ground level. whatever the initial sum chosen. to match the monkey. before the queen steps out of the bottom basket. The son steps out and the weight descends to the ground. [Carroll. the weight at the other end of the rope rises also. 9) 244. When the traveller crosses the International Date Line. the weight will move up less than the monkey. against the weight. which was internationally agreed with just such a purpose In mind. 1961. Lower the top basket. [Carroll. 269) 248. Given friction in the pulley wheel. long after the question had first troubled Carroll. p. but only in 1884. The queen gets into the basket containing the weight and descends against the weight of son and daughter. containing the weight. 1978. place the weight in it and lower the weight. 1961. 4) 246. if the pulley is sufficiently stiff. p. and the son goes down in the other basket.

Moreover. and 3 to the customer. 'A level mile takes i hour. 4 and 3. 1954. Hence in 6 hours they went 12 miles out and 12 back. 317)
[Cuthwellis. p. it makes not the slightest difference whether the water and brandy were stirred a little or a lot. 1978.The Solutions
251
249. so the customer has given the shopkeeper 7s 3d. [Hudson. 77) 251. [Carroll. or whether the glasses contained equal quantities of liquid initially. if nearly all up hill. For simplicity of calculation. up hill}. down hill t. the shopkeeper gives 60 to the customer and 12 + 1 to his friend. The amounts are equal. 14)
. and the friend has 48. thus. The quantity of liquid in each glass at the end is the same as at the start. 12 and 1. If the 12 miles out had been nearly all level. Then the customer gives 120 + 6 to the shopkeeper and 24 to the friend. and the shopkeeper is 87d better off. as they started at 3. 24 and 6. the shopkeeper has 60. whether on the level or on the hill-side. takes! an hour. Hence to go and return over the same mile. On balance the friend has gained or lost nothing. Appendix A) 250. p. and the friend gives 30 + 4 to the shopkeeper. 30. and therefore what one has lost. 1961. turn the amounts given into modern pence. p. 1958. they would have taken a little over 3 hours. a little under 4.' [Carroll. the other has gained. they got there within t an hour of t past 6. Hence 3! hours must be within! an hour of the time taken in reaching the peak. so that the customer has 120.

'But Barnum had an April Fool possibility up his sleeve. Its thirty-third leap takes it to the 99-foot mark and so another leap. 1959. But it jumps only two-thirds as quickly as the cat. The dog. 'In this remarkable problem we find that the lake contained exactly 11 acres. is compelled to go 102 feet and back.
E I---------'~
L---~7~--~----~~----~B
. carrying it two feet beyond the mark. Suppose that the cat is named Sir Thomas. the cat will have travelled only 90 feet and 8 inches. which proves that on any right-angle triangle the square o( the longest side is equal to the sum of the squares of the other two sides.
853 7491638897 5992 3969 3745 2247 2247 [Loyd. and the dog is female! The phrase "she makes three leaps to his two" would then mean that the dog would leap 9 feet while the cat went 4. No. the dog must make 68 leaps to go the distance. on the contrary.252
Penguin Book of Curious and Interesting Puzzles
253. 14]
254. Thus when the dog finishes the race in 68 leaps.' [Loyd. This definite answer is worked out by the Pythagorean law. Book 1. 41]
255. It has to make precisely 100 leaps to complete the distance and return. becomes necessary. No. In all. therefore the approximate answer of "nearly 11 acres" is not sufficiently correct. 'The cat wins. of course. Book 1. 1959. so that while the cat is making 100 leaps the dog cannot quite make 67.

1959. prove the square estate on BC to equal 116 acres. 36] 256. "Oh. No. The square of its sides. The pieces are placed to make a little white horse in the centre as shown. 102]
. 1954. AD being 9 units long and BD 17. No.5. which equals 76.5 acres.5 to prove that the lake contains exactly 11 acres. Book 1. 45] 257. we deduct the same from 76. Since the areas of the oblong and two triangles can plainly be seen to be 65. because 9 x 9 equals 81. Book 1. which added to 17 x 17 (289) equals the 370 acres of the largest field.
[Loyd. Book 1.The Solutions
253
'In the illustration ABD represents our triangle.
'It was this trick of the White Horse of Uffington which popularized the slang expression. No. The area of our triangle ADB is clearly half of 9 x 17. 4 and 10.' [Loyd. but that is a horse of another colour!'" [Loyd. 'The black pieces of paper are nothing but a delusion and a snare. 1959. CBF also is a right-angle triangle. AEC is a right-angle triangle. and the square of 5 (25) added to the square of 7 (49) shows that the square on AC equals 74.

then the cow stands + 5 from the other. 86] 260. the position is obtainable. the minimum possible.1) and 5(x . All the possible positions in which the fifteen titles and the single space might be arranged. 69] 261. a value of 48 feet. Book 1. The cow can travel (ix . but quite impossible to reach any 'even' position. Since we are told that the train travelled at 90 miles per hour. This is forty-four moves. With this information we write the equation:
2x . [Loyd. The boy is five years old. The original puzzle is impossible to solve. If we let x be the bridge's length in feet.5 from one end and !x
This gives x.1). so we see that the train is five times faster than the cow. two such swops are made. 166]
. These two periods of time reduce to (x . 1959.1) + (3x . The train is 2x from the nearest end. 1959. Book 2. can be divided into two equal classes.1). we know the cow's gait to be 18 miles per hour. Book 2. No. call them the 'odd' and the 'even' positions.1 = 5(fx .5) + (fx + 41) in the same time that the train travels (2x . the length of the bridge. however. No. In particular. If. From an 'odd' it is possible to reach any other 'odd' position. [Loyd. 21] 259. No. Five odd 'figures' will add up to 14 as follows: 11
1
1
14 [Loyd. the new position can never be obtained from the original position. The actual speed of the train plays no part whatever in this calculation. Book 2. if in any particular position one pair of tiles is swopped. No.5)
!x .254
Penguin Book of Curious and Interesting Puzzles
258. [Loyd. 1959. but we need to know it in order to learn the speed of the cow. 1959. Move the pieces in this sequence: 14 11 1287610 12874364 7 14 11 15 13 9 12 8 4 10 8 4 14 11 15 13 9 124 8 5 4 8 9 13 14 10 6 2 1.

No. The cost of this would be $1. and the two trips are completed in unit time.1 = 0. 'Let 1 be the length of the army and the time it takes the army to march its length. each trip is a distance of 1 relative to the army. No. We multiply this by 50 to get the final answer of 120. The cheapest way to make an endless chain out of the six fivelink pieces is to open up all five links of one piece.I on his forward trip. his speed relative to the moving army will be x . x + 1 on hiS backward trip.) As before. which is 20 cents cheaper than the cost of a new endless chain.2x . for which x has the positive value of 1 + j2. and on his two diagonal trips. On the courier's forward trip. Book 1. Book 2.' [Loyd.30. then use them for joining the remaining five pieces into an endless chain. 1959. and since he completes the four trips in unit time we can write:
F-=I
. Let x be the total distance travelled by the courier and also his speed. so we can write the following equation
--+--=
x-I x+l
1
1
This can be expressed as the quadratic: x 2 .The Solutions
255
262. Each trip is a distance of 1 (relative to the army). (It does not matter where he starts his round trip. 25) 263. so to simplify the problem we think of him as starting at a rear corner of the square instead of at the centre of the rear. 51) 264. '[In Part II) the courier's speed relative to the moving army is x . 1959.
[Loyd. The army's speed will also be 1.1.7 miles. On the return trip his speed relative to the army will be x + 1.

. Martin Gardner comments.04 miles while B is walking. Band C remain on the bicycle for the remainder of the journey. 1959. No. No. He and A. 'Loyd's Cyclopaedia does not answer this difficult problem . This is multiplied by 50 to get the final answer of 209.056 miles.18112. supported by the answers to similar problems in Dudeney's puzzle books. seems to be the following: 'C. A dismounts.4x' + 4x + 5 = 0. 123) 266. ride the tandem for 31..256
Penguin Book of Curious and Interesting Puzzles
--+--+
x-I x+l
1
1
Jx2=1 =
2
2Xl
This can be expressed as the fourth degree equation: X4 . The total time is a little less than 2. Book 2. The best procedure.' [Loyd. and C turns around and rides back to pick up B at a spot 5. always rides the tandem.3 hours. the fastest walker. arriving at the same time that A arrives on foot.63 miles from the start. Book 2. 1959. the slowest walker.
.' [Loyd. 146)
265. which has only one root that fits the problem's conditions: 4.

The historical incident was 'the dropping of the tea into the sea'. the first day of the week. 'The girl weighed 11 H pounds when she arrived. Kate promised to marry Danny "when the week after next is the week before last". 33] 276. her promise would have fallen due. Therefore she will marry Danny in 28 days after her promise. 1928. 29] 275. 1928.The Solutions
259
[Loyd (lnr). Had she promised a day earlier.
[Loyd (lnr). otherwise known as the Boston Tea Party. 1928. 26] 273. [Loyd (lnr).' [Loyd (lnr). p. 22 days later. which increased her weight by 10 per cent. p. p. 27] 274. then on Sunday. She ate 1~ pounds of breakfast food and gathered 10 pounds of samples. 'On Sunday. 1928. p. 1928. 47]
. p.' [Loyd (lnr).

'
. now. and then when they are closed up in one direction they form the triangle. on 17th March 1905. It will be seen that the four pieces form a sort of chain. and make 1K equal to BE. in the more general form: "A New Problem on Superposition: a demonstration that an equilateral triangle can be cut into four pieces that may be reassembled to form a square. and also at the Royal Institution in the following month. [White. Bisect AB in D and BC in E." 'I add an illustration showing the puzzle in a rather curious practical form. from the points D and K drop perpendiculars on E1 to Land M . as it was made in polished mahogany with brass hinges for use by certain audiences. bisect AF in G and describe the arc AHF. from E with distance EH. and when closed in the other direction they form the square. 58-9]
282.. pp. at Burlington House. 2 Kf2 Qf6..262
Penguin Book of Curious and Interesting Puzzles
(e) 1 f4 e5. 'The illustration will show that the triangular piece of cloth may be cut into four pieces that will fit together and form a perfect square. and EH is the length of the side of the required square. 3 Kg3 and Black can force perpetual check.
'I exhibited this problem before the Royal Society. produce the line AE to F making EF equal to EB. produce EB to H. with some examples of a general method for transforming all rectilinear figures into squares by dissection. describe the arc H1. 1913.

I show four of these ways.. 26]
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'Imagine the room to be a cardboard box.. so that the cardboard may be laid on the table...\·"·
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263
[Dudeney.
1
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2
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.. FLOOR
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t.. and indicate in every case the relative
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FI. 1907. Then the box may be cut in various ways.. I
. No.

a Dutchman cannot catch a black hog and a Dutchwoman can never capture a white one! But each can. 78] 285. so that the eighth plank can be used as a bridge to the other side. he will have no difficulty whatever in securing both in a very few moves. and it will be found that the shortest route is in No.'
[Dudeney. and that the white hog can never be caught by Katrun. The diagram shows how seven of the planks are used to 'round off' the corner.' [Dudeney. 'So if the first player just determines that he will send Hendrick after the white porker and Katrun after the black one. 'Each hog merely runs in and out of one of the nearest corners and can never be captured.4. It will be seen that the spider actually passes along five of the six sides of the room!' [Dudeney. for it is only 40 feet in length. No. 1907. without difficulty.264
Penguin Book of Curious and Interesting Puzzles
positions of the spider and the fly. Lower the piece on the right one "tooth" and they will form a perfect square. No. 'The diagram shows how the piece of bunting is to be cut into two pieces. catch one of the other colour. and the straightened course which the spider must take without going off the cardboard.
. 77] 286. These are the four most favourable cases. No. 1907. curious as it must at first sight appear. 'A very short examination of this puzzle game should convince the reader that Hendrick can never catch the black hog. The fact is. 1907. with the roses symmetrically placed. 75] 284.

-_. while standing in the bows. This is the simplest solution.266
Penguin Book of Curious and Interesting Puzzles
288. in which a dog runs back and forth between his master and an approaching friend. (During that time the fly has been continuously flying in one direction or the other. If the jester had fastened the end of his rope to the stern of the boat and then. p. had given a series of violent jerks.=_-=-:. 'No doubt some of my readers will smile at the statement that a man in a boat on smooth water can pull himself across with the tiller rope! But it is a fact. the boat would have been propelled forward.' [Dudeney. at 150 miles an hour. This has often been put to practical test. 26] 289.. Since the cars are approaching each other at a combined speed of 150 miles an hour. 1907. when the fly will be crushed. they meet after 3001150 = 2 hours. 52] 290. The fly first meets car B in 1 hour 48 minutes.) [Dudeney. though another version. This appears in 536 Puzzles and Curious Problems. The fly has therefore flown a total distance of 300 miles. 1967.
d
. and it is said that a speed of two or three miles an hour may be attained. Question (1) is inserted merely in order to induce the reader to solve question (2) by adding up a long series of fractions. by using an intermediate square. No. had appeared in Modern Puzzles.

' In this way the first player can be certain of always placing the last cigar. J. which is then dissected by the dotted lines into the Greek Cross. B. apart from itself and unity: 1. 1926. So. the four pieces will form the square in the second. or 4649 cats caught 239 mice each. This puzzle is typical of Dudeney's interest in properties of numbers. a and b. [Dudeney. to take him home. No. travels another 40 miles and has 9 gallons to take him home. [Dudeney. 3. No. 'The first player must place his first cigar on end in the exact centre of the table. whatever the second player may do throughout.' This process is repeated. 22 and 28 inches from an end. 1907. [Dudeney.The Solutions
267
The crescent is in the form of two equal straight lines. left to return home. 'The nine men.111. He also gave another solution.25 and 31 inches from one end. No.111. when B transfers 1 gall. and has 1 gall. with the marks at 1.4. No.23. 10. when A transfers 1 gall. Dudeney placed the marks at 1. G. with a total of ten pieces. 2. having travelled 360 miles out and home. 16. until finally the last man. joined by two identical circular arcs. Making the cuts in the first figure. 1917. 14. 37] 291. 180] 294. 1907. [Dudeney. 1926. both factors being prime. to each of the other eight. C. 4. 47] 292. F. [Dudeney. The eight go another 40 miles. H. 49]
. 16. ]. 398] 293. A. No.111 = 239 x 4649. the first player must always repeat it in an exactly diametrically opposite position.111 has only two factors. E. to each of the other seven and has 2 galls. Now. 1.5. D. all go 40 miles together on the 1 gall. in their engine tanks. either 239 cats caught 4649 mice each.

1917. No. and.
Penguin Book of Curious and Interesting Puzzles
[Dudeney. it is necessary to start at one of these cells and end at the other. 32] 297.
[Dudeney. with the exception of hole 5. This is a variant of the Bridges of Konigsberg. which is reached in three drives of 125 yards and one stroke backwards of 100 yards. [Dudeney. 1907. Therefore. 1907. 317] 296. in order to pass through every door just once.268
295. No. and the fair demoiselle is at the other starred cell. it is possible to go round in just twenty-six strokes: the strokes required for all the holes are evident. With a choice of 125 or 100 yards. No. There are just two cells with an odd number of doors. the route must start at the starred 'odd' cell on the outside. 34]
.

59 Ibs and 621bs. B. I
----i
. as the weight of the five trusses together.Dudeney notes that if it is larger than the square in area.The Solutions
269
298. must weigh 1211bs. 101) 299. by mere subtraction. No.I
[Dudeney. We thus know that A. The distance AB is equal to one half of the hypotenuse of the triangle. we find the weights of the other five trusses.' [Dudeney. D and E. and C and E must weigh 1201bs. 56 Ibs. Now. and the next lightest.
I
I
I
I
. the lightest being A and the heaviest E. 112 Ibs. If we call the five trusses in the order of weight. B. No. respectively. A. 1917. the weight of A and C. 'Add together the ten weights and divide by 4 and we get 289 Ibs. Then the two heaviest. D and E weigh together 2311bs. The triangle should not be too large .I
-__
. This beautiful solution requires just five pieces and only two cuts. 152)
. C. which gives us the weight of C as 581bs.I
I . D and E. 1917. 54 Ibs. then a dissection requires six pieces. then the lightest 110 Ibs must be the weight of A and B.

J.270
Penguin Book of Curious and Interesting Puzzles
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419] 302. 'According to the conditions. 23] 303. 1917. 1926. was SZ + 6 2 = 61 in the year 54 + 64 = 1921. and 2 x 31 = 62 in the year 2 x 31' = 1922.
N+--=Nx N-l
N
(1 + . other than unity. No. 93] 304. No. 1926. Jenkins was born in 1860. and 3 x 5 = 15 in the year 3 x 54 = 1875. there is no possible solution to this puzzle. in the strict sense in which one at first understands them.1) have this property. then the difficulty is got over. Three coins are placed as on the left. each touching the others. and then two coins are added as in the second figure. In such a dilemma one always has to look for some verbal quibble or trick. whatever the integral value of N. [Dudeney. Nand N/(N .The Solutions
271
301. De Morgan was born in 1806 and was 43 in the year 43' = 1849 (see problem 231).'
. No.. If the owner of house A will allow the water company to run their pipes for houses Band C through his property (and we are not bound to assume that he would object).
[Dudeney.N 1)
2
=--
N-l
N-l
[Dudeney. as shown in our illustration.

Finally slide out 5 and place it against 2 and 3. and move 3 to just touch 6 and 5. This is also exact. 1926. [Dudeney. No. or sometimes of transforming a triangle of six pennies (move 4 in the first figure to touch 5 and 6) into the circle. 1917.
This puzzle is usually presented as the problem of merely transforming the first arrangement into the final arrangement. Dudeney's presentation seems to me much superior. Next move coin 6 as in the next figure. No.272
Penguin Book of Curious and Interesting Puzzles
[Dudeney. This can be done exactly. 251) 305. Arrange the six pennies as in the first figure. 213)
.

as shown by A. the three diameters will form a right-angled triangle. will exactly equal Harry's small bun. 1917.one quarter of the confectionery each. No. Fred's piece.half the rim of the other. H. B. with the addition of the piece marked G . Therefore. they will have their fair shares . It follows that the two smaller buns are exactly equal to the large bun. 'The secret of the bun puzzle lies in the fact that. with the relative dimensions of the circles as given. 1917. 429] 307. 148]
.The Solutions
306.'
[Dudeney. and there are only five pieces necessary. if we give David and Edgar the two halves marked D and E. No. on the top of the remaining one and trace its circumference in the manner shown. Thus each boy gets an exactly equal share. H. C. Then if we place the small bun. F. Thirteen coins can be placed as shown in this figure:
273
[Dudeney.

177]
......... r .. and also the short perpendicular lines half an inch apart. cut solidly out of the cardboard. of course. -
.--· . half an inch from the edges. First rule pencil lines B Band C C..'
~r-------
________________________________ 1
. though the dimensions are of no importance.. No.274
Penguin Book of Curious and Interesting Puzzles
308...... ··... and in order that they shall coincide it is well to prick through the card with a needle the points where the short lines end. without any join. 'It is an interesting variant of the puzzle to cut out two keys on a ring .. and if you cut away all the shaded parts you will get the chain. 'The reader will probably feel rewarded for ~ny care and patience that he may bestow on cutting out the cardboard chain. We will suppose that he has a piece of cardboard measuring 8 in.. Then cut right through the card along all the short perpendicular lines. and half through the card along the short portions of B Band C C that are not dotted.. and from D D up to C C. 1917.. Rule lines on the other side in just the same way. &
D~
__________________________________________
--J~
2
[Dudeney... _
~8~'~~
____
~A
Bf--· . Now take your penknife and split the card from A A down to B B.. Yet if you want a long chain you must. take a long strip of cardboard. by 2! in. the cardboard may now be divided into two interlacing ladder-like portions. r-. -
... With a little careful separation of the parts with the penknife.without join..... Next turn the card over and cut half through along the short lines on B B and C C at the places that are immediately beneath the dotted lines on the upper side. r .

un before placing in the new position. Though originally derived from the Latin word ovum. As a matter of fact. and the idea is useful in all such cases where it is a condition that the pieces must be different in shape. 160) 310. 'One object that I had in view when presenting this little puzzle was to point out the uncertainty of the meaning conveyed by the word "oval".the Yin and Yang of the great Monad. It will be seen that the shape of the horseshoe is more easily determined from the circle than the dimensions of the circle from the horseshoe. in giving either B or D a quarter >:. 'The puzzle was to cut the two shoes (including the hoof contained within the outlines) into four pieces. that fit together and form the perfect circle shown in diagram 3. while some eggs are spherical in shape. and a sphere or circle is most
. yet what we understand as the egg-shape (with one end smaller than the other) is only one of many forms of the oval. a curvilinear square. 1917. No. It will further be observed that the two pieces A and B of one shoe and the two pieces C and D of the other form two exactly similar halves of the circle .' [Dudeney. though the latter presents no difficulty when you know that the curve of the long side of the shoe is part of the circumference of your circle. all different in shape. it is a puzzle based on the principle contained in that curious Chinese symbol the Monad. to the piece B. It was also stipulated that all four pieces should be different in shape. In forming D we simply add on a symmetrical piece. It will be noticed that 1 and 2 are cut into the required four pieces.
'The above diagrams give the correct solution to the problem.The Solutions
275
309. that would fit together and form a perfect circle. an egg. a precisely similar effect must be produced. The difference between Band D is instructive. two pieces each. Therefore.

It is correct to say that an oval is an oblong curvilinear figure. whose ignorance of the poultry-yard led him to substitute the word "hen" for "fowl".
. and this includes the ellipse. every oval is not an ellipse. If wt.we are on safer ground. remarking. but here we must be careful of error. having two unequal diameters. but all other figures which in any way approach towards the form of an oval without necessarily having the properties above described are included in the term "oval". speak of an ellipse . every hen is not a cock!" Similarly.a conical ellipse . gentlemen. and bounded by a curved line returning into itself. "We must remember.276
Penguin Book of Curious and Interesting Puzzles
certainly not an oval. Thus the following solution that I give to our puzzle involves the pointed "oval" known among architects as the "vesica piscis". we must always note that although every ellipse is an oval. that although every cock is a hen. many years ago. I recollect a Liverpool town councillor.

making only six pieces in all. [Strand Magazine.184 625 Distance = . a bit of firewood from a bundle. in competition. The sum is 9567 + 1085 = 10652. But I wished to keep the same number as in the original story. since the time in our problem is given in seconds and part of the distance in inches. Since there are 60 x 60 3600 (seconds)
. These holes are a trifle longer than those in the schoolmaster's stools. let us spread out its walls. 'The first step is to find the distance travelled by the spiders. for points.65 x 5280 x 12 41.184 (inches) 1 (hour) gives us our velocity. a piece of tin from a toy engine. It will be seen that the eight pieces form two stools of exactly the same size and shape with similar hand-holes. 'When I first gave the above puzzle in a London newspaper. They are a fairly serviceable pair of compasses.65 (mile) 0.x = 650 inches 3600 11 That is how far each spider travelled. no correct solution was received. Of course 5 and 6 can be cut out in one piece .also 7 and 8 . No. the velocity is 0. But we were asked to find the dimensions of the room. 157] 311. Calculating: 0. but we want to get this in inches per second. ceiling. Then 41. We use the formula
DIStance = Velocity x Time.65 miles per hour.
According to our stated conditions.. now. and two portions of a hairpin. the cuts being made along the other lines. much as if we were to open out the six faces of a cardboard box to make one flat piece. and floor onto a plane. a tin tack. I was compelled to improvise with the aid of a small penknife. but they are much narrower and of considerably smaller area. and I shall keep them as a memento of your puzzle" . [Dudeney.The Solutions
277
'The dotted lines in the table are given for greater clearness. July 1924] 312. but an ingenious and neatly executed attempt by a man lying in a London infirmary was accompanied by the following note: "Having no compasses here. 1917. Therefore.

c
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F
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8
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several possible paths. .
I
.278
Penguin Book of Curious and Interesting Puzzles
:Er-----.I
. ..... keeping the wall from which the spiders start fixed.
.
.
\
"
"
f
F'
.
I
\
.
. We can then see that the following eight paths are possible:
.
I
.
III
I
. this must be done in all possible ways. and laying down the others in such a way as to keep every face attached to another along a common edge (see diagram).

3.7.2)
+
w)'
= 650
inches
(3. 4. Over the ceiling and one side wall and parts of the two end walls. 8. and h = height. Denote the three dimensions of the room by I = length.4)
(h : w + 80Y+ (I + h: w .
279
Over the floor and parts of the two end walls. we have
21 + h + w = 1280
whence
1 + . w = width.stance
= J 160' +
+
h = 650 inches.
(I
(1.= 640..The Solutions 1. 2. 7.6.2)
(3. Over the floor and one side wall and parts of the two end walls. Over one side wall and parts of the two end walls.80Y= 650'
From these we find:
(I
(5.8)
= 650 inches
To eliminate the radical signs. Over the other side wall and parts of the two end walls. 5.6.900 = 630'
l+w=630
(I)
(II)
Adding (I) and (II).4)
=
J(
h+ w -2-
h + w + 80)' + ( 1 + .8)
+
1 + h = 650 w)' = 650' . Over the ceiling and parts of the two end walls.2 .160' = 396. we may rewrite these equations:
(I
160'
+
(I
+ W = 650' + w)' = 650'
(1. The lengths of the various paths may then be expressed: Distance = I Distance D. 2
h
+
w
The third equation now yields
. Over the Roor and the other side wall and parts of the two end walls.7.. Over the ceiling